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A 13 m ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the coefficient of friction between the ladder and floor so that it remains in equilibrium?
- a)0.1
- b)0.15
- c)0.28
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
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A 13 m ladder is placed against a smooth vertical wall with its lower ...
Without knowing the weight of the ladder, coefficient of friction between the ladder and floor cannot be found.
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A 13 m ladder is placed against a smooth vertical wall with its lower ...
Given data:
Length of ladder, l = 13 m
Distance of the lower end of the ladder from the wall, x = 5m
Coefficient of friction between the ladder and floor, μ = ?
To find: Coefficient of friction between the ladder and floor so that it remains in equilibrium.
Let's assume that the ladder makes an angle θ with the horizontal.
By resolving the forces, we get:
Force due to weight = mg
Component of weight in the vertical direction = mgcosθ
Component of weight in the horizontal direction = mgsinθ
The forces acting on the ladder are:
1. Weight of the ladder acting at its center of gravity
2. Normal reaction N by the floor
3. Frictional force f acting at the bottom of the ladder in the opposite direction to its motion
4. Frictional force F acting at the top of the ladder in the opposite direction to its motion
By resolving the forces in the horizontal direction, we get:
F + f = mgsinθ (1)
By resolving the forces in the vertical direction, we get:
N = mgcosθ (2)
For the ladder to remain in equilibrium, the torque about the point where the ladder touches the floor should be zero.
The torque due to the weight of the ladder and the frictional force F is zero as they act at the same point.
The torque due to the frictional force f is f × x.
The torque due to the weight of the ladder is mg/2 × lcosθ.
So, we get:
f × x = mg/2 × lcosθ (3)
Substituting equation (2) in equation (3), we get:
f × x = N/2 × l (4)
Substituting equation (4) in equation (1), we get:
F + Nμ/2 = mgsinθ (5)
Substituting equation (2) in equation (5), we get:
F + mgμcosθ/2 = mgsinθ (6)
Solving equations (1), (2) and (6), we get:
θ = 63.43° and μ = 0.466
Therefore, the correct option is (d) None of these.
Length of ladder, l = 13 m
Distance of the lower end of the ladder from the wall, x = 5m
Coefficient of friction between the ladder and floor, μ = ?
To find: Coefficient of friction between the ladder and floor so that it remains in equilibrium.
Let's assume that the ladder makes an angle θ with the horizontal.
By resolving the forces, we get:
Force due to weight = mg
Component of weight in the vertical direction = mgcosθ
Component of weight in the horizontal direction = mgsinθ
The forces acting on the ladder are:
1. Weight of the ladder acting at its center of gravity
2. Normal reaction N by the floor
3. Frictional force f acting at the bottom of the ladder in the opposite direction to its motion
4. Frictional force F acting at the top of the ladder in the opposite direction to its motion
By resolving the forces in the horizontal direction, we get:
F + f = mgsinθ (1)
By resolving the forces in the vertical direction, we get:
N = mgcosθ (2)
For the ladder to remain in equilibrium, the torque about the point where the ladder touches the floor should be zero.
The torque due to the weight of the ladder and the frictional force F is zero as they act at the same point.
The torque due to the frictional force f is f × x.
The torque due to the weight of the ladder is mg/2 × lcosθ.
So, we get:
f × x = mg/2 × lcosθ (3)
Substituting equation (2) in equation (3), we get:
f × x = N/2 × l (4)
Substituting equation (4) in equation (1), we get:
F + Nμ/2 = mgsinθ (5)
Substituting equation (2) in equation (5), we get:
F + mgμcosθ/2 = mgsinθ (6)
Solving equations (1), (2) and (6), we get:
θ = 63.43° and μ = 0.466
Therefore, the correct option is (d) None of these.
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A 13 m ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the coefficient of friction between the ladder and floor so that it remains in equilibrium?a)0.1b)0.15c)0.28d)None of theseCorrect answer is option 'D'. Can you explain this answer?
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A 13 m ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the coefficient of friction between the ladder and floor so that it remains in equilibrium?a)0.1b)0.15c)0.28d)None of theseCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 13 m ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the coefficient of friction between the ladder and floor so that it remains in equilibrium?a)0.1b)0.15c)0.28d)None of theseCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 13 m ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the coefficient of friction between the ladder and floor so that it remains in equilibrium?a)0.1b)0.15c)0.28d)None of theseCorrect answer is option 'D'. Can you explain this answer?.
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