Question

A 2m long ladder of mass 10kg is kept against a wall such that its base is 1.2m away from the wall. The wall is smooth but the ground is rough. Roughness of the ground is such that it offers a maximum horizontal resistive force (for sliding motion) half that of normal reaction at the point of contact. A monkey of mass 20kg starts climbing the ladder. How far can it climb along the ladder? How much is the horizontal reaction at the wall?​

Answer 1

Answer:

Reaction force due to wall in horizontal direction is 150 N

Monkey will reach to the distance 1.5 m from the bottom

Explanation:

When monkey reached to maximum possible distance then at that position the ladder is just in equilibrium

so here by Force balance

$F_f = N_2$

$m_1 g + m_2 g = N_1$

also we know that at this time when ladder is just in equilibrium then friction is half of normal force

$\frac{N_1}{2} = N_2$

so we have

$N_1 = (10 + 20)\times 10$

$N_1 = 300 N$

$N_2 = 150 N$

so horizontal reaction force due to wall is 150 N

now by torque balance about the contact point at the ground we can say

$m_1g (1 cos\theta) + m_2 g(L cos\theta) = N_2 (2 sin\theta)$

$(10 \times 10) + (20 \times 10) L = 150 (2) tan\theta$

here we know that

$cos\theta = \frac{1.2}{2}$

$\theta = 53.13$

$100 + 200 L = 300 \times \frac{4}{3}$

$L = 1.5 m$

Monkey will reach to the distance 1.5 m from the bottom

#Learn

Topic : Friction force and equilibrium

https://brainly.in/question/8299532

Answer 2

Explanation:

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