A 3.0 m long ladder of weight 98. N leans against a frictionless wall at an angle of 60.° with respect to the horizontal.
The coefficient of static friction between the ground and
the ladder
is µs=0.45.
If the ladder is not sliding, the magnitude of the force of static friction between the ground and the ladder is ...
To find the magnitude of the force of static friction between the ground and the ladder, we need to consider the equilibrium conditions.
1. First, let's identify the forces acting on the ladder:
- The weight of the ladder acts vertically downward and has a magnitude of 98 N.
- The normal force acts perpendicular to the ground and opposes the weight. Since the ladder is not sliding, the vertical component of the normal force cancels out the weight.
- The force of static friction acts horizontally between the ladder and the ground, opposing the tendency of the ladder to slide.
2. Next, let's resolve forces parallel to the ground and perpendicular to the ground:
- Parallel forces: The horizontal component of the weight is given by Wsinθ, where θ is the angle between the ladder and the horizontal. In this case, θ = 60°, so the horizontal component of the weight is 98 N * sin(60°) = 84.848 N.
- Perpendicular forces: The vertical component of the weight is given by Wcosθ. For a ladder leaning against a wall, this component is equal to the normal force, which cancels out the weight. Therefore, the vertical component of the weight is 98 N * cos(60°) = 49 N.
3. Now, let's analyze the equilibrium condition in the horizontal direction:
Since the ladder is not sliding, the force of static friction must balance the horizontal component of the weight:
Static friction force = Horizontal component of weight = 84.848 N.
Therefore, the magnitude of the force of static friction between the ground and the ladder is found to be 84.848 N.