a)
The torque produced by the load (τ) is given by the formula:
τ = F x r x sin(θ)
where F is the force (or load) applied, r is the distance (length of the crane's arm), and θ is the angle between the force and the lever arm.
We are given the maximum load the crane can handle (F = 465 N), the length of the arm (r = 13.1 m), and the angle between the horizontal and the arm (θ = 20°). We can now calculate the maximum torque (τ):
τ = 465 N x 13.1 m x sin(20°)
τ ≈ 1740.34 N*m
The magnitude of the maximum torque the crane can withstand is approximately 1740 N*m.
b)
We are now asked to find the maximum load (F) at an angle of 25° with the horizontal. Since we already know the maximum torque the crane can withstand (τ = 1740.34 N*m), we can rearrange the torque formula to find the force (F):
F = τ / (r x sin(θ))
Plugging in the given values (τ = 1740.34 N*m, r = 13.1 m, and θ = 25°):
F ≈ 1740.34 N*m / (13.1 m x sin(25°))
F ≈ 431.56 N
The maximum load for this crane at an angle of 25° with the horizontal is approximately 431.56 N.