Answer to Question #164468 in Physics for Cris

Question

Using the things at your home,for example a wrench, a beam or any thing that produces torque, demonstrate the following:

1. In static equilibrium, meaning that the sum of the forces and sum of the torque is zero.

2. In a non-static equilibrium, meaning that the system has the tendency to rotate.

In each of the two scenarios, show your computation of net force and net torque.

*Be creative in computing the applied force. You can use weights (F=mg), or you can do simple  approximation but make sure you add justification about the assumption.

(Have a picture or video of the demonstration)

Answer 1

Two painters carry a plank of plywood that they use for scaffolding over their heads on their way to the job site. The plank has a uniform mass distribution. On top it is a can of paint weighing one third as much as the plank. The painter in the rear is holding the plank at the very end and the painter in front is holding the plank one quarter of the the plank length from the front. The can of paint is two-fifths of the plank length from the front. Find the percentage of the total weight carried by the painter in front. Assume that the plank is horizontal as they carry it.

We start by identifying the object in equilibrium (the plank), and drawing a free-body diagram for it (we'll call the length of the plank L ). We will choose the pivot to be the back of the plank, and will refer to the weights of the can of paint and plank as W and 3W, respectively. Also we have chosen an (x,y) coordinate system and the positive direction of rotation to be clockwise, as shown in the diagram.

Next apply the conditions of equilibrium. Clearly the x -direction forces are not meaningful, and the y -direction force equation and torque equations are:

$$0=N_1-3W-W+N_2\\0=N_1(0)+3W(0.5L)+W(0.6L)-N_2(0.75L)$$

The L 's cancel out of the torque equation, resulting in a relation between the force exerted by the front painter and the weight of the can:

$$N_2=\frac{14}{5}W$$

The total weight carried by the two painters is found from the force equation (or from common sense), and equals 4W . So the percentage of the total weight carried by the front painter is:

$$\frac{N_2}{4W}=70$$