Question

001 (part 1 of 2) 10.0 points the arm of a crane at a construction site is 16.0 m long, and it makes an angle of 14.5° with the horizontal. assume that the max- imum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. what maximum torque can the crane with- stand if the maximum load the crane can handle is 620 n? answer in units of n·m. 002 (part 2 of 2) 10.0 points what is the maximum load for this crane at an angle of 30.1° with the horizontal? answer in units of n.

Explanation:

Step1: Identify torque formula

Torque is $\tau = rF\sin\theta$, where $r$ is arm length, $F$ is load force, $\theta$ is angle between arm and horizontal (since torque depends on perpendicular distance from pivot).

Step2: Solve for max torque (001)

Given $r=16.0\ \text{m}$, $F=620\ \text{N}$, $\theta=14.5^\circ$.
$\tau = 16.0 \times 620 \times \sin(14.5^\circ)$
First calculate $\sin(14.5^\circ) \approx 0.2504$
$\tau \approx 16.0 \times 620 \times 0.2504 = 16.0 \times 155.248 = 2483.968$

Step3: Solve for max load (002)

Rearrange torque formula for $F$: $F = \frac{\tau}{r\sin\theta}$
Use $\tau=2483.968\ \text{N·m}$, $r=16.0\ \text{m}$, $\theta=30.1^\circ$.
$\sin(30.1^\circ) \approx 0.5015$
$F \approx \frac{2483.968}{16.0 \times 0.5015} = \frac{2483.968}{8.024} \approx 309.57$

Answer:

001: $2480\ \text{N·m}$ (rounded to reasonable precision)
002: $310\ \text{N}$ (rounded to reasonable precision)