A 257 kg piano sits 1/4 of the way from the end of a 116 kg uniform, horizontal beam that has support at each end. What is the vertical force on the support furthest from the piano?

The problem involves understanding static equilibrium of forces to calculate the vertical force on the furthest support when a piano is placed 1/4 of the way on a beam. The force on the right support will be smaller due to the greater separation from the piano.

Explanation:

This question is about the concept of static equilibrium. In this scenario, the piano acts as a downwards force applied at a distance 1/4 of the length of the beam away from its end. We will use the principle that the sum of moments (or torques) about any point is zero, since the beam isn't turning or moving.

Let's denote the distance from left support to the piano as L1 (1/4 of the beam length), the force of the left support as F1, the distance of the right support to the piano as L2 (3/4 of the beam length), and the force of the right support as F2.

Using the equilibrium conditions, we can say that F1 + F2 = the weight of beam + the weight of piano, and that F1*L1 = F2*L2. By substituting the known values into these equations, we can solve for F1 and F2.

According to these calculations, the vertical force on the support furthest from the piano (F2) will be smaller than F1, since it is further from the piano.

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