A plank 3m long and having a mass of 20kg is supported at its ends. Imagine that a person with a mass of 50 kg stands m from one end. What are the magnitudes of the force exerted by the supports?
To find the magnitudes of the forces exerted by the supports on the plank, we will need to consider the torques acting on the plank.
First, let's calculate the center of mass of the plank. The center of mass of a uniform plank is at the midpoint. Since the plank is 3m long, the center of mass is at 1.5m from either end.
Next, we need to calculate the torques. Torque is the product of the force applied and the distance from the pivot point. In this case, the pivot point is where each support exerts force.
Let's consider the torque about the left support (the one closest to the person). The force exerted by the support can be denoted as F1. The person exerts a force of Fp = mg (where m is the person's mass and g is the acceleration due to gravity) at a distance of m from the left support. The weight of the plank also exerts a force of Fw = mg at its center of mass, which is 1.5m away from the left support.
To ensure equilibrium, the sum of the torques must be zero (since the plank is not rotating). Therefore, the torque exerted by the person must balance the torque exerted by the weight of the plank.
T1 = T2
Fp * m = Fw * (3 - 1.5)
Substituting the known values:
mg * m = mg * (3 - 1.5)
50 kg * 9.8 m/s^2 * m = 20 kg * 9.8 m/s^2 * (3 - 1.5)
Simplifying:
490 mN = 294 mN
Therefore, the force exerted by the left support (F1) is 294 N, and the force exerted by the right support (F2) is also 294 N since both supports are evenly supporting the plank.
the supports each exert 98 N to support the plank
one support exerts [490 N * m / 3]
to support the person
the other support exerts
... 490 N - [490 N * m / 3]