What is the minimum coeffecient of static friction ? min required between the la

Question

What is the minimum coeffecient of static friction ?min required between the ladder and the ground so that the ladder does not slip?. Express umin min in terms of m1, m2, d, L, and ?.

B.Suppose that the actual coefficent of friction is one and a half times as large as the value of

?min. That is, ?s=(3/2)?min. Under these circumstances, what is the magnitude of the force of friction f that the floor applies to the ladder?

Express your answer in terms of m1, m2, d, L, g, and ?. Remember to pay attention to the relation of force and ?s.

Comment

Explanation / Answer

Balancing vertical forces,
N2 = m1 + m2

frictional force from right to left at the bottom of the ladder with the floor,
F = u_min * N2 = u_min * (m1 + m2)

Taking moments about the point of contact of the ladder with the wall,
N2 * Lcos? - F * Lsin? - m1 * (L - d) cos? - m2 * (L/2) cos? = 0
(m1+m2) * Lcos? - u_min * (m1+m2) * Lsin? - m1 * (L - d) cos? - m2 * (L/2) cos? = 0
u_min * (m1+m2) * Lsin? = (m1+m2) * Lcos? - m1 * (L - d) cos? - m2 * (L/2) cos?
u_min * (m1+m2) * Lsin? = m1 * dcos? + (m2/2) Lcos?

u_min = (dm1 + Lm2/2)cot? / [L(m1+m2)]

then u_min is
u_min = [(d/L)m1 + (1/2)m2] cot? / (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 Lsin? + m2g (L/2)cos? + m1g dcos? = 0

F = N1 = (m2/2 + m1d/L) gcot?

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