An 70-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth, frictionless wall. If the ladder has a mass of 30 kg and it makes an angle of 60° with the ground, find the force of friction of the ground on the foot of the ladder.
To solve this problem, we need to break down the forces acting on the ladder.
First, let's consider the forces acting on the ladder itself:
1. Weight of the ladder (30 kg x 9.8 m/s^2) = 294 N, directed downwards.
2. Normal force exerted by the wall on the ladder: Since the wall is smooth and frictionless, the only vertical force acting on the ladder is the component of the weight perpendicular to the wall. This component can be calculated using trigonometry: (30 kg x 9.8 m/s^2) * cos(60°) = 147 N, directed perpendicular to the wall.
3. Tension force exerted by the man on the ladder: Since the man is not accelerating vertically, the net vertical force on the man and the ladder system must be zero. Therefore, the tension force balances the component of the weight parallel to the ladder. This component can be calculated using trigonometry: (30 kg x 9.8 m/s^2) * sin(60°) = 255 N, directed upwards.
Now, let's consider the forces acting on the man:
1. Weight of the man (70 kg x 9.8 m/s^2) = 686 N, directed downwards.
2. Tension force exerted by the man on the ladder: Since the man is not accelerating horizontally, the net horizontal force on the man must be zero. Therefore, the tension force balances the force of friction at the foot of the ladder.
Since the man is one-fourth of the way up the ladder, the distance from the man's position to the foot of the ladder is (1/4) * 10 m = 2.5 m.
To determine the force of friction at the foot of the ladder, we need to analyze the rotational equilibrium about the foot of the ladder:
1. Torque due to the weight of the ladder = (2.5 m) * (30 kg x 9.8 m/s^2) * cos(60°) = 367.5 Nm, clockwise.
2. Torque due to the tension force exerted by the man = (2.5 m) * (70 kg x 9.8 m/s^2) = 1715 Nm, counterclockwise (since the force is directed upwards).
3. Torque due to the force of friction at the foot of the ladder = 0 (since the foot of the ladder is the pivot point).
Since we are considering rotational equilibrium, the sum of the torques must be zero:
Torque due to the weight + Torque due to the tension force = 0
367.5 Nm - 1715 Nm = 0
-1347.5 Nm = 0
Therefore, the force of friction at the foot of the ladder is zero.
In conclusion, the force of friction of the ground on the foot of the ladder is zero.