The torque is $\tau = F \cdot r \, sin(\theta),$ where θ is the angle between the rotating object and the applied force.

(a) The torque is

$$ \begin{align} \tau_a &= F \cdot r \, sin(\theta) \\[5pt] &= 50 \, N (1.2 \, m) \cancel{sin(90˚)} \\[5pt] &= 60 \, Nm \end{align}$$

Notice that $sin(90˚) = 1,$ so when the applied force is at a right angle to the target, we can ignore the $sin(\theta)$ term.

(b) The torque is

$$ \begin{align} \tau_b &= F \cdot r \, sin(\theta) \\[5pt] &= 50 \, N (1.2 \, m) sin(70˚) \\[5pt] &= 56.4 \, Nm \end{align}$$

Changing the angle of the force by 20˚ from perpendicular reduces the torque.

(c) The torque is

$$ \begin{align} \tau_c &= F \cdot r \, sin(\theta) \\[5pt] &= 50 \, N (1.2 \, m) sin(110˚) \\[5pt] &= 56.4 \, Nm \end{align}$$

Notice that the torques in (b) and (c) are the same. Applying the 50N force at 70˚ and 110˚ both result in a force that is 20˚ off of perpendicular (θ = 90˚), just on different sides of it. The result is the same torque.

The torque on the left side is

$$ \begin{align} \tau &= F \, r \, sin(\theta) \\[5pt] &= 200 \, N (7 \, m) sin(45˚) \\[5pt] &= 989.9 \, N m \end{align}$$

Now if the system is in equilibrium, then the torque on the right must be the same as on the left, so the force (tension) in the wire is

$$ \begin{align} F &= \frac{\tau}{r \, sin(\theta)} \\[5pt] &= \frac{989.9 \, N \cancel{m}}{5 \cancel{m} \, sin(30˚)} \\[5pt] &= 396 \, N \end{align}$$

More force is required in the shorter wire because the distance to the pivot point is shorter, and because the angle of attachment is larger than 45˚

The minimum force will be one that is at a 90˚ angle to the wrench. Any other angle will require more force. To calculate the force, rearrange $\tau = F \cdot r$ to

$$F = \frac{\tau}{r}$$

$$ \begin{align} F &= \frac{42.0 \, N\cancel{m}}{0.35 \cancel{m}} \\[5pt] &= {\bf 14.7 \, N} \end{align}$$

Notice that if we kept the $sin(\theta)$ term, we'd get

$$ \begin{align} \tau &= F \, r \, sin(\theta) \\[5pt] F &= \frac{\tau}{r \cdot sin(\theta)} \end{align}$$

As θ diverges from 90˚, the value of sin(θ) decreases either way you go, so as the denominator of this equation decreases, the force increases.

Here is a picture of what's going on in this problem.

If the person (F₃) weren't standing on the platform, it would be fine, but the weight of the window washer will change the center of mass, and thush the pivot point. We have to first recalculate the center of mass (com) of the platform, including the window washer. It is located at:

$$ \begin{align} com &= \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} \\[5pt] &= \frac{\left( \frac{150}{9.81} \right) (2.5) + 60(3.8)}{15.3 + 60} \\[5pt] &= 3.54 \, m \end{align}$$

That's the distance to the center of mass from the left side. The corrected distances from the center of mass (the pivot point) look like this:

The ropes suspend the entire weight of the platform plus the window washer. That's

$$W = 150 \, N + 60(9.81) \, N = 738.6 \, N$$

Now the sum of all torques on the system must be zero because it's in equilibrium, so that gives us:

$$ \begin{align} F_1(3.54) + 588.6(0.26)& \\[5pt] - (738.6 - F_1)(1.46) &= 0 \\[5pt] 3.54 F_1 + 1.46 F_1 + 153 - 1078 &= 0 \\[5pt] 5.0 F_1 &= 925 \\[5pt] F_1 &= 185 \, N \end{align}$$

Then it follows that $F_2 = 554 \, N.$