A 75 kg window cleaner uses a 10 kg ladder that is 7.2 m long. He places one end on the ground 4.5 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 5.2 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?
To solve this problem, we can use the principles of torque and equilibrium.
Let's start by finding the force on the window from the ladder.
(a) The force on the window from the ladder can be determined by considering the torque acting on the ladder about the point where it rests on the ground. This torque can be balanced by the torque due to the weight of the window cleaner.
The torque due to the weight of the window cleaner is given by:
Torque = (weight of window cleaner) * (distance from the point of contact to the line of action of the weight)
Weight of window cleaner = mass * gravitational acceleration
= 75 kg * 9.8 m/s^2
= 735 N (Newtons)
The distance from the point of contact to the line of action of the weight can be calculated using similar triangles. Since the window cleaner is 5.2 m up along the ladder when the window breaks, the distance from the point of contact to the line of action of the weight is (7.2 - 5.2) m.
Distance = 7.2 m - 5.2 m
= 2 m
Therefore, the torque due to the weight of the window cleaner is:
Torque = 735 N * 2 m
= 1470 Nm (Newton-meters)
For equilibrium, the torque due to the weight of the window cleaner must be balanced by the torque due to the force on the window from the ladder.
Torque due to the force on the window from the ladder = (Force on the window) * (distance from the point of contact to the line of action of the force)
Since the ladder rests against a cracked window, the distance from the point of contact to the line of action of the force is the length of the ladder, which is 7.2 m.
Therefore, to balance the torques, we have:
(Torque due to the force on the window from the ladder) = (Torque due to the weight of the window cleaner)
(Force on the window) * (7.2 m) = 1470 Nm
Solving for the force on the window from the ladder:
Force on the window = 1470 Nm / 7.2 m
= 204.17 N (approximately)
Therefore, the magnitude of the force on the window from the ladder is approximately 204.17 N.
(b) Now, let's find the magnitude of the force on the ladder from the ground.
Since the ladder is in equilibrium, the sum of the forces in the vertical direction must be zero. Therefore, the magnitude of the force on the ladder from the ground is equal to the weight of the window cleaner.
Magnitude of the force on the ladder from the ground = weight of window cleaner
= mass * gravitational acceleration
= 75 kg * 9.8 m/s^2
= 735 N
Therefore, the magnitude of the force on the ladder from the ground is 735 N.
(c) Finally, let's find the angle (relative to the horizontal) of that force on the ladder.
To find the angle, we can use trigonometry. The force on the ladder from the ground and the length of the ladder form a right-angled triangle.
The cosine of the angle is given by the ratio of the adjacent side (force on the ladder) to the hypotenuse (length of the ladder).
Cosine(angle) = (Magnitude of the force on the ladder from the ground) / (Length of the ladder)
Cosine(angle) = 735 N / 7.2 m
Solving for the angle:
angle = arccos(735 N / 7.2 m)
= 88.49 degrees (approximately)
Therefore, the angle (relative to the horizontal) of the force on the ladder is approximately 88.49 degrees.