A uniform ladder of length 8 m and mass 12 kg stands on a rough horizontal surface inclined at an angle \( 30^{\circ} \) relative to the ground> A bunch of banana of mass 6 kg is resting on the ladder at a distance 6 m from the ground along the ladder. A 60 kg monkey notices tha banana and starts climbimg up the ladder to eat the bbanana. How far up the ladder from the ground will the monkey climb before the ladder begins to slip given that the coefficient of static friction is 0.3

To solve this problem, we need to analyze the forces acting on the ladder and the conditions for static equilibrium. Here are the steps we'll follow: 1. **Identify the forces acting on the ladder:** - Weight of the ladder (\( W_L \)): \( W_L = m_L \cdot g \) - Weight of the bananas (\( W_B \)): \( W_B = m_B \cdot g \) - Weight of the monkey (\( W_M \)): \( W_M = m_M \cdot g \) - Normal force (\( N \)) acting vertically upward at the base of the ladder. - Frictional force (\( F_f \)) acting horizontally at the base of the ladder. 2. **Calculate the weights:** - Mass of the ladder (\( m_L = 12 \, \text{kg} \)) - Mass of the bananas (\( m_B = 6 \, \text{kg} \)) - Mass of the monkey (\( m_M = 60 \, \text{kg} \)) - Acceleration due to gravity (\( g = 9.81 \, \text{m/s}^2 \)) 3. **Set up the equilibrium conditions:** - The sum of vertical forces must equal zero. - The sum of horizontal forces must equal zero. - The sum of moments about the base of the ladder must equal zero. 4. **Determine the maximum static friction force:** - The maximum static friction force is given by \( F_f = \mu_s \cdot N \), where \( \mu_s = 0.3 \). 5. **Calculate the distance the monkey can climb before the ladder slips.** Let's perform the calculations step by step. ### Step 1: Calculate the weights \[ W_L = m_L \cdot g = 12 \cdot 9.81 \] \[ W_B = m_B \cdot g = 6 \cdot 9.81 \] \[ W_M = m_M \cdot g = 60 \cdot 9.81 \] ### Step 2: Calculate the total weight and normal force The total weight acting downwards is: \[ W_{total} = W_L + W_B + W_M \] The normal force \( N \) is equal to the total weight acting vertically downwards. ### Step 3: Set up the equilibrium conditions 1. **Vertical forces:** \[ N = W_{total} \] 2. **Horizontal forces:** The frictional force must balance the horizontal component of the weight of the ladder: \[ F_f = \mu_s \cdot N \] 3. **Moments about the base of the ladder:** The moments due to the weights of the ladder, bananas, and monkey must balance the moment due to the frictional force. Now, let's calculate the weights and the total weight. I'll perform these calculations now. Calculate the value by following steps: - step0: Calculate: \(6\times 9.81\) - step1: Multiply the numbers: \(58.86\) Calculate or simplify the expression \( 12 * 9.81 \). Calculate the value by following steps: - step0: Calculate: \(12\times 9.81\) - step1: Multiply the numbers: \(117.72\) Calculate or simplify the expression \( 60 * 9.81 \). Calculate the value by following steps: - step0: Calculate: \(60\times 9.81\) - step1: Multiply the numbers: \(588.6\) ### Step 1: Calculate the weights 1. **Weight of the ladder (\( W_L \))**: \[ W_L = 12 \cdot 9.81 = 117.72 \, \text{N} \] 2. **Weight of the bananas (\( W_B \))**: \[ W_B = 6 \cdot 9.81 = 58.86 \, \text{N} \] 3. **Weight of the monkey (\( W_M \))**: \[ W_M = 60 \cdot 9.81 = 588.6 \, \text{N} \] ### Step 2: Calculate the total weight and normal force The total weight acting downwards is: \[ W_{total} = W_L + W_B + W_M = 117.72 + 58.86 + 588.6 = 765.18 \, \text{N} \] The normal force \( N \) is equal to the total weight: \[ N = W_{total} = 765.18 \, \text{N} \] ### Step 3: Calculate the maximum static friction force The maximum static friction force is given by: \[ F_f = \mu_s \cdot N = 0.3 \cdot 765.18 \] Now, let's calculate \( F_f \). Calculate the value by following steps: - step0: Calculate: \(0.3\times 765.18\) - step1: Multiply the numbers: \(229.554\) ### Step 3: Calculate the maximum static friction force The maximum static friction force is: \[ F_f = 0.3 \cdot 765.18 = 229.554 \, \text{N} \] ### Step 4: Set up the moment equilibrium condition To find out how far the monkey can climb before the ladder begins to slip, we need to consider the moments about the base of the ladder. The moments due to the weights of the ladder, bananas, and monkey must balance the moment due to the frictional force. 1. **Distance from the base to the center of mass of the ladder**: The center of mass of the ladder is at \( \frac{8}{2} = 4 \, \text{m} \) from the base. 2. **Distance from the base to the bananas**: The bananas are located at \( 6 \, \text{m} \) along the ladder. 3. **Distance from the base to the monkey**: Let \( x \) be the distance the monkey climbs up the ladder. The moments about the base of the ladder can be expressed as follows: - Moment due to the weight of the ladder: \[ M_L = W_L \cdot (4 \cdot \cos(30^\circ)) \] - Moment due to the weight of the bananas: \[ M_B = W_B \cdot (6 \cdot \cos(30^\circ)) \] - Moment due to the weight of the monkey: \[ M_M = W_M \cdot (x \cdot \cos(30^\circ)) \] - Moment due to the frictional force: \[ M_f = F_f \cdot (8 \cdot \sin(30^\circ)) \] Setting the sum of moments about the base equal to zero gives us: \[ M_L + M_B + M_M = M_f \] Now, let's calculate the moments and set up the equation. ### Step 5: Calculate the moments 1. **Calculate \( \cos(30^\circ) \) and \( \sin(30^\circ) \)**: \[ \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(30^\circ) = \frac{1}{2} \] 2. **Calculate the moments**: - Moment due to the ladder: \[ M_L = 117.72 \cdot (4 \cdot \frac{\sqrt{3}}{2}) \] - Moment due to the bananas: \[ M_B = 58.86 \cdot (6 \cdot \frac{\sqrt{3}}{2}) \] - Moment due to the monkey: \[ M_M = 588.6 \cdot (x \cdot \frac{\sqrt{3}}{2}) \] - Moment due to the frictional force: \[ M_f = 229.554 \cdot (8 \cdot \frac{1}{2}) \] Now, let's calculate these moments. Calculate the value by following steps: - step0: Calculate: \(229.554\left(8\times \frac{1}{2}\right)\) - step1: Remove the parentheses: \(229.554\times 8\times \frac{1}{2}\) - step2: Multiply the terms: \(1836.432\times \frac{1}{2}\) - step3: Convert the expressions: \(\frac{229554}{125}\times \frac{1}{2}\) - step4: Multiply the numbers: \(\frac{114777}{125}\) Calculate or simplify the expression \( 58.86 * (6 * (\sqrt(3)/2)) \). Calculate the value by following steps: - step0: Calculate: \(58.86\left(6\left(\frac{\sqrt{3}}{2}\right)\right)\) - step1: Remove the parentheses: \(58.86\times 6\left(\frac{\sqrt{3}}{2}\right)\) - step2: Remove the parentheses: \(58.86\times 6\times \frac{\sqrt{3}}{2}\) - step3: Multiply the terms: \(353.16\times \frac{\sqrt{3}}{2}\) - step4: Convert the expressions: \(\frac{8829}{25}\times \frac{\sqrt{3}}{2}\) - step5: Multiply the numbers: \(\frac{8829\sqrt{3}}{50}\) Calculate or simplify the expression \( 117.72 * (4 * (\sqrt(3)/2)) \). Calculate the value by following steps: - step0: Calculate: \(117.72\left(4\left(\frac{\sqrt{3}}{2}\right)\right)\) - step1: Remove the parentheses: \(117.72\times 4\left(\frac{\sqrt{3}}{2}\right)\) - step2: Remove the parentheses: \(117.72\times 4\times \frac{\sqrt{3}}{2}\) - step3: Multiply the terms: \(470.88\times \frac{\sqrt{3}}{2}\) - step4: Convert the expressions: \(\frac{11772}{25}\times \frac{\sqrt{3}}{2}\) - step5: Multiply the numbers: \(\frac{5886\sqrt{3}}{25}\) ### Step 5: Calculate the moments 1. **Moment due to the ladder (\( M_L \))**: \[ M_L = 117.72 \cdot (4 \cdot \frac{\sqrt{3}}{2}) \approx 407.79 \, \text{N m} \] 2. **Moment due to the bananas (\( M_B \))**: \[ M_B = 58.86 \cdot (6 \cdot \frac{\sqrt{3}}{2}) \approx 305.85 \, \text{N m} \] 3. **Moment due to the frictional force (\( M_f \))**: \[ M_f = 229.554 \cdot (8 \cdot \frac{1}{2}) \approx 918.22 \, \text{N m} \] ### Step 6: Set up the equilibrium equation Now we can set up the equation for moments about the base of the ladder: \[ M_L + M_B + M_M = M_f \] Substituting the values we have: \[ 407.79 + 305.85 + 588.6 \cdot \left( x \cdot \frac{\sqrt{3}}{2} \right) = 918.22 \] ### Step 7: Solve for \( x \) Rearranging the equation gives: \[ 588.6 \cdot \left( x \cdot \frac{\sqrt{3}}{2} \right) = 918.22 - 407.79 - 305.85 \] Calculating the right side: \[ 918.22 - 407.79 - 305.85 = 204.58 \] Now we can solve for \( x \): \[ 588.6 \cdot \left( x \cdot \frac{\sqrt{3}}{2} \right) = 204.58 \] Let's isolate \( x \): \[ x = \frac{204.58}{588.6 \cdot \frac{\sqrt{3}}{2}} \] Now, let's calculate \( x \). Calculate the value by following steps: - step0: Calculate: \(\frac{204.58}{\left(588.6\left(\frac{\sqrt{3}}{2}\right)\right)}\) - step1: Remove the parentheses: \(\frac{204.58}{588.6\left(\frac{\sqrt{3}}{2}\right)}\) - step2: Remove the parentheses: \(\frac{204.58}{588.6\times \frac{\sqrt{3}}{2}}\) - step3: Multiply the numbers: \(\frac{204.58}{\frac{2943\sqrt{3}}{10}}\) - step4: Convert the expressions: \(\frac{\frac{10229}{50}}{\frac{2943\sqrt{3}}{10}}\) - step5: Multiply by the reciprocal: \(\frac{10229}{50}\times \frac{10}{2943\sqrt{3}}\) - step6: Reduce the numbers: \(\frac{10229}{5}\times \frac{1}{2943\sqrt{3}}\) - step7: Multiply the fractions: \(\frac{10229}{5\times 2943\sqrt{3}}\) - step8: Multiply: \(\frac{10229}{14715\sqrt{3}}\) - step9: Multiply by the Conjugate: \(\frac{10229\sqrt{3}}{14715\sqrt{3}\times \sqrt{3}}\) - step10: Multiply the numbers: \(\frac{10229\sqrt{3}}{44145}\) ### Step 7: Solve for \( x \) The distance \( x \) that the monkey can climb up the ladder before it begins to slip is approximately: \[ x \approx 0.40134 \, \text{m} \] ### Conclusion The monkey can climb approximately **0.40 meters** up the ladder from the ground before the ladder begins to slip.