A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2, and the other end is connected to box 3. The weights of the three boxes are W1 = 50.7 N, W2 = 32.1 N, and W3 = 28.6 N. Determine the magnitude of the normal force that the table exerts on box 1.
367.5
To determine the magnitude of the normal force that the table exerts on box 1, we need to analyze the forces acting on box 1.
First, let's consider the forces acting on box 1. We have the weight of box 1 (W1 = 50.7 N) acting downward. The normal force is exerted by the table and acts perpendicular to the table surface. Since the table is in contact with box 1, the normal force acts upward.
Since the rope is massless, the tension in the rope is equal on both sides of the pulley. Therefore, the tension transmitted to box 1 is the same as the tension transmitted to box 2. Let's call this tension T.
Considering box 2, we have the weight of box 2 (W2 = 32.1 N) acting downward and the tension T acting upward.
Applying Newton's second law to box 1 in the vertical direction, we have:
∑Fy = -W1 + N = -ma
Where:
∑Fy is the net force in the y-direction (vertical direction)
N is the normal force
W1 is the weight of box 1
m is the mass of box 1 (which we need to find)
a is the acceleration
Since the rope is inextensible, the acceleration of both boxes is the same. Therefore, we can set the acceleration, a, equal to zero.
- W1 + N = 0
Rearranging the equation, we have:
N = W1
So, the magnitude of the normal force that the table exerts on box 1 is equal to the weight of box 1, which is 50.7 N.