| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we need to apply the concept of equilibrium, which means that the net force and the net torque on the system must be zero. Let's denote the mass of Throckmorton as m. The center of mass of the plank is at the midpoint, which is 3.0 m from each end. Since the sawhorses are equidistant from the center of the plank, each sawhorse supports half of the weight of the plank. The weight of the plank is Mg, where g is the acceleration due to gravity (approximately 9.8 m/s²). The force exerted by each sawhorse on the plank is Mg/2. Since Throckmorton is standing on the right-hand end of the plank, the force he exerts is mg, and it acts at a distance of 3.0 m from the left sawhorse and 1.5 m from the right sawhorse. To ensure the plank remains at rest, the net torque about any point must be zero. Let's consider the torque about the left sawhorse. The torque due to the weight of the plank is (Mg/2)(1.5 m) clockwise, and the torque due to Throckmorton's weight is (mg)(3.0 m) counterclockwise. Setting the net torque equal to zero, we have: (Mg/2)(1.5 m) = (mg)(3.0 m) Simplifying the equation: (90 kg)(9.8 m/s²)(1.5 m) / (3.0 m) = mg Solving for m: m = (90 kg)(9.8 m/s²)(1.5 m) / (3.0 m)(9.8 m/s²) m = 45 kg Therefore, Throckmorton can be a maximum of 45 kg for the plank to remain at rest. |
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| deepseekr1 | To determine the maximum mass Throckmorton can be without tipping the plank, we analyze the torque equilibrium around the right sawhorse. The plank is 6.0 meters long with a mass of 90 kg, and the sawhorses are 1.5 meters apart, equidistant from the center. The right sawhorse is 3.75 meters from the left end, and Throckmorton stands on the right end (6.0 meters from the left end). 1. **Torque due to the plank's weight**: The center of mass of the plank is at 3.0 meters, which is 0.75 meters to the left of the right sawhorse. The torque created by the plank's weight is: \[ \text{Torque}_{\text{plank}} = 90 \, \text{kg} \times g \times 0.75 \, \text{m} \] 2. **Torque due to Throckmorton's weight**: Throckmorton stands 2.25 meters to the right of the right sawhorse. The torque created by his weight is: \[ \text{Torque}_{\text{Throckmorton}} = m \times g \times 2.25 \, \text{m} \] 3. **Equilibrium condition**: For the plank to remain in equilibrium, the torques must balance: \[ 90 \, \text{kg} \times 0.75 \, \text{m} = m \times 2.25 \, \text{m} \] Solving for \( m \): \[ 90 \times 0.75 = m \times 2.25 \] \[ 67.5 = 2.25m \] \[ m = \frac{67.5}{2.25} = 30 \, \text{kg} \] Thus, the maximum mass Throckmorton can be is \boxed{30} kg. |
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| gpto1 | To solve this problem, we'll analyze the plank in static equilibrium under the action of gravitational forces and the supports (sawhorses). The key is to determine the maximum mass \( M_T \) that Cousin Throckmorton can have without causing the plank to tip over. **Given:** - Length of plank, \( L = 6.0 \, \text{m} \) - Mass of the plank, \( M_{\text{plank}} = 90 \, \text{kg} \) - Distance between supports, \( D = 1.5 \, \text{m} \) - Supports are equidistant from the center, so each is \( d = \frac{D}{2} = 0.75 \, \text{m} \) from the center. - Throckmorton stands at the right end of the plank (at \( x = +3.0 \, \text{m} \) from the center). **Steps:** 1. **Diagram Setup:** Let's establish a coordinate system with the center of the plank at \( x = 0 \). The plank extends from \( x = -3.0 \, \text{m} \) to \( x = +3.0 \, \text{m} \). - Left support at \( x_A = -0.75 \, \text{m} \) - Right support at \( x_B = +0.75 \, \text{m} \) - Center of mass of the plank at \( x_{\text{cm}} = 0 \, \text{m} \) - Throckmorton at \( x_T = +3.0 \, \text{m} \) 2. **Identify the Tipping Point:** The plank will start to tip about the left support at \( x_A \) when Throckmorton's weight causes the normal force at the right support to become zero. At this point, the only pivot is at \( x_A \). 3. **Calculate Lever Arms:** - Distance from pivot to plank's center of mass: \[ r_{\text{plank}} = x_{\text{cm}} - x_A = 0 - (-0.75) = 0.75 \, \text{m} \] - Distance from pivot to Throckmorton: \[ r_T = x_T - x_A = (+3.0) - (-0.75) = 3.75 \, \text{m} \] 4. **Apply the Torque Equilibrium Condition:** For rotational equilibrium about point \( x_A \): \[ \sum \tau = 0 \] The torques due to the weights of the plank and Throckmorton are: - Torque due to the plank (counterclockwise): \[ \tau_{\text{plank}} = -M_{\text{plank}} \cdot g \cdot r_{\text{plank}} \] - Torque due to Throckmorton (clockwise): \[ \tau_T = +M_T \cdot g \cdot r_T \] Setting the net torque to zero: \[ \tau_T + \tau_{\text{plank}} = 0 \] Simplify: \[ M_T \cdot g \cdot r_T - M_{\text{plank}} \cdot g \cdot r_{\text{plank}} = 0 \] Cancel \( g \) (acceleration due to gravity) from both sides: \[ M_T \cdot r_T = M_{\text{plank}} \cdot r_{\text{plank}} \] 5. **Solve for Throckmorton's Mass \( M_T \):** \[ M_T = \frac{M_{\text{plank}} \cdot r_{\text{plank}}}{r_T} \] Substitute the known values: \[ M_T = \frac{90 \, \text{kg} \times 0.75 \, \text{m}}{3.75 \, \text{m}} \] \[ M_T = \frac{67.5 \, \text{kg}\cdot\text{m}}{3.75 \, \text{m}} \] \[ M_T = 18 \, \text{kg} \] **Answer:** The maximum mass Throckmorton can have is **18 kilograms** without causing the plank to tip over. **Result:** \[ M_T = 18 \, \text{kg} \] |
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| GPT-4o | To solve this problem, we need to analyze the torques acting on the plank to ensure it remains in static equilibrium when Cousin Throckmorton stands on it. Let's denote the mass of Cousin Throckmorton as \( m_T \). The plank is supported by two sawhorses located symmetrically about its center. Thus, each sawhorse is \( D/2 = 1.5/2 = 0.75 \, \text{m} \) from the center. Key points: - Length of plank \( L = 6.0 \, \text{m} \) - Mass of plank \( M = 90 \, \text{kg} \) - Distance between sawhorses \( D = 1.5 \, \text{m} \) - Distance of each sawhorse from center \( d = 0.75 \, \text{m} \) - Position of Throckmorton at the right end of the plank ### Step-by-step Solution: 1. **Determine the positions of the sawhorses:** - The left sawhorse is at \( x = 3.0 - 0.75 = 2.25 \, \text{m} \) - The right sawhorse is at \( x = 3.0 + 0.75 = 3.75 \, \text{m} \) 2. **Calculate the force exerted by the plank's own weight:** - The weight of the plank acts at its center \( x = 3.0 \, \text{m} \) - The gravitational force due to the plank is \( F_g = Mg = 90 \times 9.8 = 882 \, \text{N} \) 3. **Torque calculation:** - Choose the left sawhorse as the pivot point for calculating torques. - For the plank to be in equilibrium, the sum of the clockwise torques must equal the sum of the counterclockwise torques. **Counterclockwise torque (due to the plank’s weight):** - Torque due to the plank's weight: \( \tau_{\text{plank}} = 882 \, \text{N} \times (3.0 - 2.25) = 882 \times 0.75 = 661.5 \, \text{N} \cdot \text{m} \) **Clockwise torque (due to Throckmorton):** - Torque due to Throckmorton: \( \tau_{\text{Throckmorton}} = m_T \times g \times (6.0 - 2.25) = m_T \times 9.8 \times 3.75 \, \text{m} \) 4. **Set the torques equal to solve for \( m_T \):** \[ 661.5 = m_T \times 9.8 \times 3.75 \] \[ m_T = \frac{661.5}{9.8 \times 3.75} \] \[ m_T = \frac{661.5}{36.75} \] \[ m_T = 18 \, \text{kg} \] ### Conclusion: The maximum mass Cousin Throckmorton can be for the plank to remain at rest is \( \boxed{18 \, \text{kg}} \). |
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| llama3.1-70B-Instruct | To solve this problem, we need to consider the torque and rotational equilibrium of the plank. Since the plank is at rest, the net torque acting on it must be zero. Let's denote the mass of Cousin Throckmorton as m. The weight of the plank (Mg) acts downward at its center, and the weight of Throckmorton (mg) acts downward at the right-hand end of the plank. The sawhorses exert upward forces on the plank, which we'll call F1 and F2. Since the plank is uniform, its weight acts at its center, which is 3.0 m from either end (L/2 = 6.0 m / 2 = 3.0 m). The distance from the center of the plank to the right-hand end is also 3.0 m. The torque due to the weight of the plank is: τ_plank = -Mg × 0 = 0 (since the weight acts at the center, the torque is zero) The torque due to the weight of Throckmorton is: τ_Throck = -mg × 3.0 m (since the weight acts at the right-hand end, 3.0 m from the center) The torque due to the sawhorses is: τ_sawhorses = F1 × 0.75 m + F2 × 0.75 m (since the sawhorses are 1.5 m apart, and the plank is 3.0 m long, the distance from the center to each sawhorse is 0.75 m) Since the plank is at rest, the net torque must be zero: τ_net = τ_plank + τ_Throck + τ_sawhorses = 0 Substituting the expressions for each torque, we get: 0 = -mg × 3.0 m + F1 × 0.75 m + F2 × 0.75 m Since the plank is in equilibrium, the sum of the forces acting on it must also be zero. This means that the upward forces exerted by the sawhorses must balance the downward weight of the plank and Throckmorton: F1 + F2 = Mg + mg Now, we can solve for m. Rearranging the equation for the net torque, we get: mg × 3.0 m = F1 × 0.75 m + F2 × 0.75 m Substituting the expression for F1 + F2, we get: mg × 3.0 m = (Mg + mg) × 0.75 m Simplifying and rearranging, we get: m = (Mg × 0.75 m) / (3.0 m × g - 0.75 m × g) Substituting the values given in the problem, we get: m = (90 kg × 9.8 m/s² × 0.75 m) / (3.0 m × 9.8 m/s² - 0.75 m × 9.8 m/s²) m ≈ 45 kg Therefore, Cousin Throckmorton can be at most 45 kg massive for the plank to remain at rest. |
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| llama3.1-405B-Instruct | To solve this problem, we need to use the concept of torque and rotational equilibrium. Since the plank is at rest, the net torque acting on it must be zero. Let's consider the forces acting on the plank: 1. The weight of the plank (W_p) acts downward at the center of the plank. 2. The weight of Throckmorton (W_T) acts downward at the right-hand end of the plank. 3. The normal forces (N_1 and N_2) exerted by the sawhorses act upward at the points where the plank rests on them. Since the plank is uniform, the weight of the plank can be considered to act at the center of the plank, which is equidistant from the sawhorses. Therefore, the torque due to the weight of the plank is zero. The torque due to the weight of Throckmorton is: τ_T = W_T \* (L/2) where L/2 is the distance from the center of the plank to the right-hand end. The torque due to the normal forces is: τ_N = N_1 \* (D/2) + N_2 \* (D/2) Since the plank is at rest, the net torque is zero: τ_T - τ_N = 0 Substituting the expressions for τ_T and τ_N, we get: W_T \* (L/2) - N_1 \* (D/2) - N_2 \* (D/2) = 0 Since the plank is uniform, the normal forces are equal (N_1 = N_2 = N). Therefore: W_T \* (L/2) - 2N \* (D/2) = 0 Simplifying and rearranging, we get: W_T = (2N \* D) / L Since the weight of Throckmorton is equal to his mass (m_T) times the acceleration due to gravity (g), we can substitute: m_T \* g = (2N \* D) / L To find the maximum mass of Throckmorton, we need to find the maximum normal force (N) that the sawhorses can exert. Since the plank is at rest, the sum of the forces acting on it must be zero: W_p + W_T - 2N = 0 Substituting the expressions for W_p and W_T, we get: M \* g + m_T \* g - 2N = 0 Simplifying and rearranging, we get: 2N = (M + m_T) \* g Substituting this expression for 2N into the previous equation, we get: m_T \* g = ((M + m_T) \* g \* D) / L Simplifying and rearranging, we get: m_T = (M \* D) / (L - D) Substituting the given values, we get: m_T = (90 kg \* 1.5 m) / (6.0 m - 1.5 m) m_T = 135 kg / 4.5 m m_T = 30 kg Therefore, the maximum mass of Throckmorton is 30 kg. |
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| DeepSeek-Coder-V2 | To solve this problem, we need to consider the conditions for static equilibrium, which are that the sum of the forces and the sum of the torques acting on the plank must both be zero. Given: - Length of the plank, L = 6.0 m - Mass of the plank, M = 90 kg - Distance between sawhorses, D = 1.5 m - Throckmorton stands on the right-hand end of the plank First, let's determine the positions of the sawhorses. Since they are equidistant from the center of the plank, each sawhorse is located at a distance of D/2 = 0.75 m from the center. Let's denote the mass of Throckmorton as m. The forces acting on the plank are: 1. The weight of the plank, Mg, acting downwards at the center of the plank. 2. The weight of Throckmorton, mg, acting downwards at the right-hand end of the plank. 3. The normal forces from the sawhorses, F1 and F2, acting upwards. Since the plank is in static equilibrium, the sum of the forces must be zero: F1 + F2 = Mg + mg Now, let's consider the torques acting on the plank. We can choose the pivot point at one of the sawhorses, say the left one. The sum of the torques around this pivot point must be zero: Torque due to the weight of the plank: τ_plank = Mg * (L/2 - D/2) = 90 * 9.81 * (3 - 0.75) = 90 * 9.81 * 2.25 = 1984.05 Nm Torque due to the weight of Throckmorton: τ_Throckmorton = mg * (L - D/2) = m * 9.81 * (6 - 0.75) = m * 9.81 * 5.25 = 51.4975m Nm Torque due to the normal force from the right sawhorse: τ_F2 = F2 * D = F2 * 1.5 Since the sum of the torques must be zero: 1984.05 + 51.4975m - F2 * 1.5 = 0 We also know that F2 = Mg + mg - F1. Since the sawhorses are equidistant from the center, F1 = F2. Therefore: F2 = (Mg + mg) / 2 = (90 * 9.81 + m * 9.81) / 2 = (882.9 + 9.81m) / 2 = 441.45 + 4.905m Substitute F2 into the torque equation: 1984.05 + 51.4975m - (441.45 + 4.905m) * 1.5 = 0 1984.05 + 51.4975m - 662.175 - 7.3575m = 0 1321.875 + 44.14m = 0 44.14m = -1321.875 m = -1321.875 / 44.14 ≈ -30 kg Since mass cannot be negative, Throckmorton can be as massive as he wants without causing the plank to rotate. However, in reality, the plank's material strength and the sawhorses' support capacity would limit Throckmorton's mass. For the purpose of this problem, we can conclude that Throckmorton can be infinitely massive while maintaining static equilibrium. |
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| Qwen2-72B-Instruct | To solve this problem, we need to apply the concept of equilibrium, which means that the net force and the net torque on the system must be zero. Let's denote the mass of Throckmorton as m. The center of mass of the plank is at the midpoint, which is 3.0 m from each end. Since the sawhorses are equidistant from the center of the plank, each sawhorse supports half of the weight of the plank. The weight of the plank is Mg, where g is the acceleration due to gravity (approximately 9.8 m/s²). The force exerted by each sawhorse on the plank is Mg/2. Since Throckmorton is standing on the right-hand end of the plank, the force he exerts is mg, and it acts at a distance of 3.0 m from the left sawhorse and 1.5 m from the right sawhorse. To ensure the plank remains at rest, the net torque about any point must be zero. Let's consider the torque about the left sawhorse. The torque due to the weight of the plank is (Mg/2)(1.5 m) clockwise, and the torque due to Throckmorton's weight is (mg)(3.0 m) counterclockwise. Setting the net torque equal to zero, we have: (Mg/2)(1.5 m) = (mg)(3.0 m) Simplifying the equation: (90 kg)(9.8 m/s²)(1.5 m) / (3.0 m) = mg Solving for m: m = (90 kg)(9.8 m/s²)(1.5 m) / (3.0 m)(9.8 m/s²) m = 45 kg Therefore, Throckmorton can be a maximum of 45 kg for the plank to remain at rest. |
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| baichuan4 | To solve this problem, we need to consider the torque and the moment of inertia of the plank with respect to its center. The plank will remain at rest if the total torque from the weight of the plank and Throckmorton is zero. First, let's calculate the torque due to the weight of the plank alone. The center of mass of the plank is at its center, which is 3.0 m from either end. The torque due to the plank's weight around the center is given by: Torque_plank = Weight_plank * Distance_from_center Weight_plank = M * g = 90 kg * 9.81 m/s^2 = 882.9 N Distance_from_center = 3.0 m Torque_plank = 882.9 N * 3.0 m = 2648.7 N*m (clockwise) Now, let's consider the torque when Throckmorton stands on the right-hand end of the plank. We need to find out how much Throckmorton can weigh such that the total torque is zero. Let's call Throckmorton's mass M_Throckmorton and his weight W_Throckmorton. Torque_Throckmorton = W_Throckmorton * Distance_from_center_Throckmorton W_Throckmorton = M_Throckmorton * g Distance_from_center_Throckmorton = 3.0 m (since Throckmorton is standing at the end, which is 3.0 m from the center) For the plank to remain at rest, the total torque must be zero: Torque_plank + Torque_Throckmorton = 0 2648.7 N*m (clockwise) + (M_Throckmorton * 9.81 m/s^2) * 3.0 m (counterclockwise) = 0 Solving for M_Throckmorton: M_Throckmorton = -2648.7 N*m / (9.81 m/s^2 * 3.0 m) M_Throckmorton = -88.2 kg The negative sign indicates that the torque due to Throckmorton's weight must counteract the torque from the plank's weight. Therefore, Throckmorton can have a maximum mass of 88.2 kg for the plank to remain at rest. |
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| doubao | To keep the plank at rest, the total torque about the center of the plank must be zero. The weight of the plank acts at its center and creates no torque. The torque due to Throckmorton's weight (mg) is m * g * (L/2) (where m is Throckmorton's mass and g is the acceleration due to gravity). The torque due to the support on the left side is Mg * (D/2) and the torque due to the support on the right side is Mg * ((L - D)/2). Setting the total torque to zero gives: m * g * (L/2) = Mg * (D/2) + Mg * ((L - D)/2) m * g * (L/2) = Mg * (L/2) m = M = 90 kg. So Throckmorton can be 90 kg. |
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| minimax | To solve this problem, we need to consider the forces acting on the plank and the conditions for the plank to remain at rest. The plank is supported by two sawhorses, and the center of mass of the plank must be directly above the line joining the two supports for the plank to be balanced. Let's denote the mass of Cousin Throckmorton as \( m \). The plank's center of mass is at its geometric center, which is at a distance \( \frac{L}{2} \) from each end. Since the sawhorses are equidistant from the center of the plank, they are each \( \frac{L}{4} \) from the end of the plank. When Throckmorton stands on the right-hand end of the plank, the plank will rotate about the left sawhorse if his weight is too great. To keep the plank at rest, the torque due to Throckmorton's weight must be balanced by the torque due to the plank's weight. The torque (\( \tau \)) is given by the product of the force (weight in this case) and the lever arm (distance from the pivot point). The lever arm for Throckmorton is \( D \), and the lever arm for the plank's weight is \( \frac{L}{2} - \frac{D}{2} \). The torque due to Throckmorton's weight is: \[ \tau_{Throckmorton} = m \cdot g \cdot D \] The torque due to the plank's weight is: \[ \tau_{plank} = M \cdot g \cdot \left(\frac{L}{2} - \frac{D}{2}\right) \] For the plank to remain at rest, these two torques must be equal and opposite: \[ m \cdot g \cdot D = M \cdot g \cdot \left(\frac{L}{2} - \frac{D}{2}\right) \] We can cancel \( g \) from both sides since it is a common factor and does not affect the mass calculation: \[ m \cdot D = M \cdot \left(\frac{L}{2} - \frac{D}{2}\right) \] Now we can solve for \( m \): \[ m = M \cdot \frac{\left(\frac{L}{2} - \frac{D}{2}\right)}{D} \] Plugging in the given values: \[ m = 90 \, \text{kg} \cdot \frac{\left(\frac{6.0 \, \text{m}}{2} - \frac{1.5 \, \text{m}}{2}\right)}{1.5 \, \text{m}} \] \[ m = 90 \, \text{kg} \cdot \frac{\left(3.0 \, \text{m} - 0.75 \, \text{m}\right)}{1.5 \, \text{m}} \] \[ m = 90 \, \text{kg} \cdot \frac{2.25 \, \text{m}}{1.5 \, \text{m}} \] \[ m = 90 \, \text{kg} \cdot 1.5 \] \[ m = 135 \, \text{kg} \] So, Cousin Throckmorton can be as massive as 135 kg for the plank to remain at rest with him standing on the right-hand end. |
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| kimi | To solve this problem, we need to consider the torques acting on the plank. The torque is the force that causes an object to rotate around an axis. In this case, the plank will rotate around the point where it is supported by the sawhorses. The torque (τ) is calculated by the formula: \[ \tau = r \times F \] where \( r \) is the distance from the pivot point (the center of the plank in this case) to the point where the force is applied, and \( F \) is the force applied. Since the plank is uniform and the sawhorses are equidistant from the center, the center of mass of the plank is at the midpoint of the plank, which is \( \frac{L}{2} = \frac{6.0 \text{ m}}{2} = 3.0 \text{ m} \) from each end. The torque due to the weight of the plank (Mg) is: \[ \tau_{\text{plank}} = r_{\text{plank}} \times F_{\text{plank}} \] \[ \tau_{\text{plank}} = 3.0 \text{ m} \times (M \times g) \] \[ \tau_{\text{plank}} = 3.0 \text{ m} \times (90 \text{ kg} \times 9.8 \text{ m/s}^2) \] \[ \tau_{\text{plank}} = 3.0 \text{ m} \times 882 \text{ kg} \cdot \text{m/s}^2 \] \[ \tau_{\text{plank}} = 2646 \text{ kg} \cdot \text{m}^2/\text{s}^2 \] Now, let's consider the torque due to Cousin Throckmorton's weight. If Throckmorton stands on the right-hand end of the plank, the distance from the center of the plank to Throckmorton is \( L - r_{\text{plank}} = 6.0 \text{ m} - 3.0 \text{ m} = 3.0 \text{ m} \). Let \( m \) be the mass of Throckmorton. The torque due to Throckmorton's weight is: \[ \tau_{\text{Throckmorton}} = r_{\text{Throckmorton}} \times F_{\text{Throckmorton}} \] \[ \tau_{\text{Throckmorton}} = 3.0 \text{ m} \times (m \times g) \] \[ \tau_{\text{Throckmorton}} = 3.0 \text{ m} \times (m \times 9.8 \text{ m/s}^2) \] For the plank to remain at rest, the torque due to Throckmorton must be equal to the torque due to the plank's weight: \[ \tau_{\text{Throckmorton}} = \tau_{\text{plank}} \] \[ 3.0 \text{ m} \times (m \times 9.8 \text{ m/s}^2) = 2646 \text{ kg} \cdot \text{m}^2/\text{s}^2 \] \[ m \times 9.8 \text{ m/s}^2 = \frac{2646 \text{ kg} \cdot \text{m}^2/\text{s}^2}{3.0 \text{ m}} \] \[ m \times 9.8 \text{ m/s}^2 = 882 \text{ kg} \cdot \text{m}^2/\text{s}^2 \] \[ m = \frac{882 \text{ kg} \cdot \text{m}^2/\text{s}^2}{9.8 \text{ m/s}^2} \] \[ m = 90 \text{ kg} \] So, Cousin Throckmorton can be a maximum of 90 kg for the plank to remain at rest. |
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| wenxinyiyan | To solve this problem, we need to consider the torques acting on the plank. The torque is the product of the force and the perpendicular distance from the force to the pivot point (in this case, the sawhorse). Let's label the sawhorses as points A and B, with A being the left sawhorse and B being the right sawhorse. The plank rests on these two points, and they are equidistant from the center of the plank. The mass of the plank, M, is uniformly distributed along its length, L. Therefore, the center of mass of the plank is at its geometric center, which is L/2 = 3.0 m from each end. Cousin Throckmorton wants to stand on the right-hand end of the plank. Let's denote his mass as m. For the plank to remain at rest, the net torque acting on it must be zero. We can calculate the torques with respect to either sawhorse A or B; the result should be the same due to symmetry. Let's choose sawhorse B as our pivot point. The torque due to the weight of the plank (acting at its center of mass) is: τ_plank = M * g * (L/2 - D) where g is the acceleration due to gravity (approximately 9.81 m/s²), L/2 is the distance from the pivot point to the center of mass of the plank, and D is the distance from the pivot point to the sawhorse (which is also the distance from the center of the plank to each sawhorse). The torque due to the weight of Cousin Throckmorton (acting at the right-hand end of the plank) is: τ_Throckmorton = m * g * (L - D) where L - D is the distance from the pivot point to the right-hand end of the plank. For the net torque to be zero, we must have: τ_plank = τ_Throckmorton M * g * (L/2 - D) = m * g * (L - D) We can now solve for m: m = (M * (L/2 - D)) / (L - D) Plugging in the given values: M = 90 kg L = 6.0 m D = 1.5 m g = 9.81 m/s² (although this will cancel out since it appears on both sides of the equation) We get: m = (90 * (6.0/2 - 1.5)) / (6.0 - 1.5) m = (90 * (3.0 - 1.5)) / 4.5 m = (90 * 1.5) / 4.5 m = 30 kg So, Cousin Throckmorton can have a maximum mass of 30 kg if the plank is to remain at rest. |
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| zhipu | To determine the maximum mass that Cousin Throckmorton can have while standing on the right-hand end of the plank without causing it to tip, we need to consider the conditions for static equilibrium. Specifically, we need to ensure that the sum of the torques (moments) around the pivot point (one of the sawhorses) is zero and that the sum of the vertical forces is zero. ### Step-by-Step Solution: 1. **Identify the Pivot Point:** We will choose the left sawhorse as the pivot point because it simplifies the calculation. The plank will start to tip if the torque due to Throckmorton's weight exceeds the torque due to the weight of the plank. 2. **Calculate the Torque Due to the Plank:** The weight of the plank \( M \) acts at its center of mass, which is at \( L/2 = 6.0 \, \text{m}/2 = 3.0 \, \text{m} \) from either end. The distance from the center of mass to the left sawhorse is \( 1.5 \, \text{m} \) (since the sawhorses are equidistant from the center). The torque due to the plank's weight about the left sawhorse is: \[ \tau_{\text{plank}} = M \cdot g \cdot \left( \frac{L}{2} - \frac{D}{2} \right) \] where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Substituting the given values: \[ \tau_{\text{plank}} = 90 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot \left( 3.0 \, \text{m} - 0.75 \, \text{m} \right) \] \[ \tau_{\text{plank}} = 90 \cdot 9.8 \cdot 2.25 \] \[ \tau_{\text{plank}} = 90 \cdot 22.05 \] \[ \tau_{\text{plank}} = 1984.5 \, \text{N} \cdot \text{m} \] 3. **Calculate the Torque Due to Throckmorton:** Throckmorton stands at the right-hand end of the plank, which is \( L = 6.0 \, \text{m} \) from the left sawhorse. The torque due to Throckmorton's weight about the left sawhorse is: \[ \tau_{\text{Throckmorton}} = m \cdot g \cdot L \] where \( m \) is Throckmorton's mass. 4. **Set Up the Equilibrium Condition:** For the plank to remain at rest, the torque due to Throckmorton must not exceed the torque due to the plank: \[ \tau_{\text{Throckmorton}} \leq \tau_{\text{plank}} \] Substituting the expressions for the torques: \[ m \cdot g \cdot L \leq 1984.5 \, \text{N} \cdot \text{m} \] \[ m \cdot 9.8 \cdot 6.0 \leq 1984.5 \] \[ m \cdot 58.8 \leq 1984.5 \] \[ m \leq \frac{1984.5}{58.8} \] \[ m \leq 33.75 \, \text{kg} \] ### Conclusion: The maximum mass that Cousin Throckmorton can have while standing on the right-hand end of the plank without causing it to tip is **33.75 kg**. |
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| gpto1 | To solve this problem, we'll analyze the plank in static equilibrium under the action of gravitational forces and the supports (sawhorses). The key is to determine the maximum mass \( M_T \) that Cousin Throckmorton can have without causing the plank to tip over. **Given:** - Length of plank, \( L = 6.0 \, \text{m} \) - Mass of the plank, \( M_{\text{plank}} = 90 \, \text{kg} \) - Distance between supports, \( D = 1.5 \, \text{m} \) - Supports are equidistant from the center, so each is \( d = \frac{D}{2} = 0.75 \, \text{m} \) from the center. - Throckmorton stands at the right end of the plank (at \( x = +3.0 \, \text{m} \) from the center). **Steps:** 1. **Diagram Setup:** Let's establish a coordinate system with the center of the plank at \( x = 0 \). The plank extends from \( x = -3.0 \, \text{m} \) to \( x = +3.0 \, \text{m} \). - Left support at \( x_A = -0.75 \, \text{m} \) - Right support at \( x_B = +0.75 \, \text{m} \) - Center of mass of the plank at \( x_{\text{cm}} = 0 \, \text{m} \) - Throckmorton at \( x_T = +3.0 \, \text{m} \) 2. **Identify the Tipping Point:** The plank will start to tip about the left support at \( x_A \) when Throckmorton's weight causes the normal force at the right support to become zero. At this point, the only pivot is at \( x_A \). 3. **Calculate Lever Arms:** - Distance from pivot to plank's center of mass: \[ r_{\text{plank}} = x_{\text{cm}} - x_A = 0 - (-0.75) = 0.75 \, \text{m} \] - Distance from pivot to Throckmorton: \[ r_T = x_T - x_A = (+3.0) - (-0.75) = 3.75 \, \text{m} \] 4. **Apply the Torque Equilibrium Condition:** For rotational equilibrium about point \( x_A \): \[ \sum \tau = 0 \] The torques due to the weights of the plank and Throckmorton are: - Torque due to the plank (counterclockwise): \[ \tau_{\text{plank}} = -M_{\text{plank}} \cdot g \cdot r_{\text{plank}} \] - Torque due to Throckmorton (clockwise): \[ \tau_T = +M_T \cdot g \cdot r_T \] Setting the net torque to zero: \[ \tau_T + \tau_{\text{plank}} = 0 \] Simplify: \[ M_T \cdot g \cdot r_T - M_{\text{plank}} \cdot g \cdot r_{\text{plank}} = 0 \] Cancel \( g \) (acceleration due to gravity) from both sides: \[ M_T \cdot r_T = M_{\text{plank}} \cdot r_{\text{plank}} \] 5. **Solve for Throckmorton's Mass \( M_T \):** \[ M_T = \frac{M_{\text{plank}} \cdot r_{\text{plank}}}{r_T} \] Substitute the known values: \[ M_T = \frac{90 \, \text{kg} \times 0.75 \, \text{m}}{3.75 \, \text{m}} \] \[ M_T = \frac{67.5 \, \text{kg}\cdot\text{m}}{3.75 \, \text{m}} \] \[ M_T = 18 \, \text{kg} \] **Answer:** The maximum mass Throckmorton can have is **18 kilograms** without causing the plank to tip over. **Result:** \[ M_T = 18 \, \text{kg} \] |
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