A 50 kg uniform ladder, 5.0 m long, is placed against a smooth wall at a height of h = 3.7 m. The base of the ladder rests on a rough horizontal surface whose coefficient of static friction is 0.40. An 80 kg block is suspended from the top rung of the ladder, just at the wall. In Figure 10.12, the magnitude of the force exerted on the base of the ladder, due to contact with the rough horizontal surface is closest to:

Question

A 50 kg uniform ladder, 5.0 m long, is placed against a smooth wall at a height of h = 3.7 m. The base of the ladder rests on a rough horizontal surface whose coefficient of static friction is 0.40. An 80 kg block is suspended from the top rung of the ladder, just at the wall. In Figure 10.12, the magnitude of the force exerted on the base of the ladder, due to contact with the rough horizontal surface is closest to:

Answer 1

To solve this problem, we need to consider the forces acting on the ladder and the equilibrium conditions.

Let's break down the forces acting on the ladder:

1. Weight of the ladder: The weight acts downward from the center of mass of the ladder and is given by the formula: F_ladder = m_ladder * g, where m_ladder is the mass of the ladder and g is the acceleration due to gravity.

2. Normal force from the horizontal surface: The normal force acts perpendicular to the surface and balances the weight of the ladder. It is equal in magnitude but opposite in direction to the weight of the ladder. N_ladder = F_ladder.

3. Friction force from the horizontal surface: The friction force opposes the motion of the ladder and acts parallel to the surface. It can be calculated using the formula: F_friction = µ * N_ladder, where µ is the coefficient of static friction.

Now, let's consider the forces acting on the ladder due to the block:

1. Tension force in the ladder: The weight of the block creates a tension force in the ladder, acting upward. Tension = m_block * g.

Since the ladder is in equilibrium (not accelerating), the sum of the forces in both the x and y directions should be zero.

In the y-direction (vertical direction):

N_ladder - F_ladder = 0

Since N_ladder and F_ladder are equal, we can combine these two forces into:

2N_ladder = m_ladder * g

In the x-direction (horizontal direction):

F_friction - Tension = 0

Substituting the values we have:

µ * N_ladder - m_block * g = 0

Since N_ladder = F_ladder, we can substitute values and solve for F_friction:

µ * F_ladder - m_block * g = 0

µ * (m_ladder * g) - m_block * g = 0

µ * m_ladder = m_block

F_ladder = m_block / µ

F_ladder = (80 kg) / (0.40)

F_ladder = 200 N

Therefore, the magnitude of the force exerted on the base of the ladder, due to contact with the rough horizontal surface, is closest to 200 Newtons.