A uniform ladder 10.0 m long is leaning against a frictionless wall at an angle of 60.0° above the horizontal. The weight of the ladder is 21.8 lb. A 57.8-lb boy climbs 4.38 m up the ladder. What is the magnitude of the frictional force exerted on the ladder by the floor?
70mtr
To find the magnitude of the frictional force exerted on the ladder by the floor, we can use the principles of equilibrium.
First, let's analyze the forces acting on the ladder. There are three forces in play: the weight of the ladder (W_ladder), the weight of the boy (W_boy), and the frictional force exerted by the floor (F_friction). The ladder is in equilibrium, which means the sum of the forces in the vertical and horizontal directions is zero.
In the vertical direction:
ΣF_vertical = F_friction + W_ladder + W_boy = 0
Since the ladder is leaning against a frictionless wall, we know that there is no vertical component of the frictional force. Thus, F_friction and W_ladder cancel each other out.
W_ladder = -F_friction (Note: The negative sign indicates that the forces are in opposite directions)
Therefore, we have:
W_boy = F_friction + W_ladder
Next, let's convert the given weights from pounds to Newtons (1 lb ≈ 4.45 N):
W_ladder = 21.8 lb × 4.45 N/lb = 96.91 N
W_boy = 57.8 lb × 4.45 N/lb = 257.51 N
Now we can substitute the values into the equation:
257.51 N = F_friction + 96.91 N
Rearranging the equation to solve for the frictional force:
F_friction = 257.51 N - 96.91 N
F_friction = 160.60 N
Therefore, the magnitude of the frictional force exerted on the ladder by the floor is approximately 160.60 Newtons.