A window washer of \(M=75 \mathrm{~kg}\) is taking a lunch break and sitting \(4 / 3 \mathrm{~m}\) from the left edge of a 5- meter-long board that hangs from two cables. The mass \(m\) of the board is \(20 \mathrm{~kg}\). If \(T_1\) is the tension in the right cable and \(T_2\) is the tension in the left cable, what are their values? (A) \(T_1=200 \mathrm{~N} ; T_2=650 \mathrm{~N}\) (B) \(T_1=300 \mathrm{~N} ; T_2=650 \mathrm{~N}\) (C) \(T_1=650 \mathrm{~N} ; T_2=300 \mathrm{~N}\) (D) \(T_1=650 \mathrm{~N} ; T_2=200 \mathrm{~N}\) (E) \(T_1=450 \mathrm{~N} ; T_2=400 \mathrm{~N}\)

Understanding torque is crucial when dealing with rotational equilibrium. Torque, often denoted by \( \tau \), is the rotational equivalent of force. It describes how much a force acting on an object causes that object to rotate. The equation for torque is \( \tau = r \times F \), where \( r \) is the distance from the pivot point to where the force is applied, and \( F \) is the force.

Torque also depends on the angle between the force and the direction from the pivot to the point of force application, though in most introductory problems like the one in our exercise, this angle is 90 degrees, simplifying the calculation to \( \tau = rF \).

As exhibited in the exercise, when calculating the torques caused by the window washer and the board, one must consider the distance from the pivot point (left edge of the board in this case) to where each weight acts. The board's weight acts directly at the point of suspension, so it creates no torque there, while the washer's weight produces a torque that tends to rotate the board clockwise.

Cables holding up structures in equilibrium are subject to tension forces. These are the forces exerted by a string, rope, cable, or similar object pulling on another object.

The tension doesn't just randomly appear; rather, it is a reaction to the forces acting on the system. In the example of the window washer, gravitational forces act on the board, and the cables must generate enough tension to counteract this force. If multiple cables are involved, as in our problem, the tension in each cable will depend on their relative positions to the weights and the pivot point.

It's important to note that tension assumes a cable is massless and unstretchable, ensuring that tension is uniform throughout its length. This simplification is often applied in introductory physics problems and is valid for the given exercise, allowing us to solve for the tensions easily.

The equilibrium condition is essential for solving physics problems involving resting or constant velocity scenarios. In such cases, all the forces are balanced, and there is no net force causing acceleration. There are two types of equilibrium to consider: translational equilibrium, where all forces sum to zero, and rotational equilibrium, where all torques sum to zero.

In the exercise, we apply the principle of rotational equilibrium where \( \sum \tau = 0 \). The window washer and board system doesn't rotate despite the various forces acting on it, indicating that the clockwise and anticlockwise torques are balancing out.

In a similar vein, translational equilibrium helps us determine that the upward forces (tensions) and the downward forces (weights) are equal, hence the equation \( T_1 + T_2 = W_1 + W_2 \). Through careful application of both the translational and rotational equilibrium conditions, we can solve for unknown forces such as tensions in cables.