Question 321.
Refer to the figure as above.
A uniform ladder with mass m₂ and length L rests against a smooth wall. A do-it-yourself enthusiast of mass m₁ stands on the ladder at a distance d from the bottom (measured along the ladder). The ladder makes an angle θ with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude ƒ between the floor and the ladder. N₁ is the magnitude of the normal force exerted by the wall on the ladder, and N₂ is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.
1)
What is the minimum coeffecient of static friction μ_min required between the ladder and the ground so that the ladder does not slip? Express μ_min in terms of m₂, m₁, d , L , and θ.
2)
Suppose that the actual coefficent of friction is one and a half times as large as the value of μ_min. That is, μ_s= (3/2) μ_min. Under these circumstances, what is the magnitude of the force of friction ƒ that the floor applies to the ladder? Express your answer in terms of m₁, m₂, d , L ,g and θ.
Answer 321.
1)
Balancing vertical forces,
N₂ = m₁ + m₂
When the ladder is on the verge of slipping, frictional force from right to left at the bottom of the ladder with the floor,
F = μ_min * N₂ = μ_min * (m₁ + m₂)
Taking moments about the point of contact of the ladder with the wall,
N₂ * Lcosθ - F * Lsinθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ = 0
=> (m₁+m₂) * Lcosθ - μ_min * (m₁+m₂) * Lsinθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ = 0
=> μ_min * (m₁+m₂) * Lsinθ = (m₁+m₂) * Lcosθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ
=> μ_min * (m₁+m₂) * Lsinθ = m₁ * dcosθ + (m₂/2) Lcosθ
=> μ_min
= (dm₁ + Lm₂/2)cotθ / [L(m₁+m₂)]
= [(d/L)m₁ + (1/2)m₂] cotθ / (m₁+m₂).
2)
Balancing vertical forces,
N₂ = m₁ + m₂
Balancing horizontal forces,
f = N₁
Taking moments of all the forces about the point of contact of the ladder with the floor,
- N₁ Lsinθ + m₂g (L/2)cosθ + m₁g dcosθ = 0
=> f = N₁ = (m₂/2 + m₁d/L) gcotθ.
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Refer to the figure as above.
A uniform ladder with mass m₂ and length L rests against a smooth wall. A do-it-yourself enthusiast of mass m₁ stands on the ladder at a distance d from the bottom (measured along the ladder). The ladder makes an angle θ with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude ƒ between the floor and the ladder. N₁ is the magnitude of the normal force exerted by the wall on the ladder, and N₂ is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.
1)
What is the minimum coeffecient of static friction μ_min required between the ladder and the ground so that the ladder does not slip? Express μ_min in terms of m₂, m₁, d , L , and θ.
2)
Suppose that the actual coefficent of friction is one and a half times as large as the value of μ_min. That is, μ_s= (3/2) μ_min. Under these circumstances, what is the magnitude of the force of friction ƒ that the floor applies to the ladder? Express your answer in terms of m₁, m₂, d , L ,g and θ.
Answer 321.
1)
Balancing vertical forces,
N₂ = m₁ + m₂
When the ladder is on the verge of slipping, frictional force from right to left at the bottom of the ladder with the floor,
F = μ_min * N₂ = μ_min * (m₁ + m₂)
Taking moments about the point of contact of the ladder with the wall,
N₂ * Lcosθ - F * Lsinθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ = 0
=> (m₁+m₂) * Lcosθ - μ_min * (m₁+m₂) * Lsinθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ = 0
=> μ_min * (m₁+m₂) * Lsinθ = (m₁+m₂) * Lcosθ - m₁ * (L - d) cosθ - m₂ * (L/2) cosθ
=> μ_min * (m₁+m₂) * Lsinθ = m₁ * dcosθ + (m₂/2) Lcosθ
=> μ_min
= (dm₁ + Lm₂/2)cotθ / [L(m₁+m₂)]
= [(d/L)m₁ + (1/2)m₂] cotθ / (m₁+m₂).
2)
Balancing vertical forces,
N₂ = m₁ + m₂
Balancing horizontal forces,
f = N₁
Taking moments of all the forces about the point of contact of the ladder with the floor,
- N₁ Lsinθ + m₂g (L/2)cosθ + m₁g dcosθ = 0
=> f = N₁ = (m₂/2 + m₁d/L) gcotθ.
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