Question
The drawing shows box 1 resting on a table, with box 2 resting on top of box 1. 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 Wy = 55 N, W2
= 35 N, and W3 = 28 N. Determine
the magnitude of the normal force that the table exerts on box 1.
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 Wy = 55 N, W2
= 35 N, and W3 = 28 N. Determine
the magnitude of the normal force that the table exerts on box 1.
Answers
GPT 3.5
To solve this problem, we need to consider the forces acting on each box.
For box 1:
- The weight of box 1 is acting downward with magnitude W1 = 55 N.
- The normal force exerted by the table on box 1 is acting upward and is what we need to find.
For box 2:
- The weight of box 2 is acting downward with magnitude W2 = 35 N.
- The tension T in the rope is acting upward.
For box 3:
- The weight of box 3 is acting downward with magnitude W3 = 28 N.
- The tension T in the rope is acting downward.
Since the rope is massless and frictionless, the tension T is the same for boxes 2 and 3.
Now, we can set up equations of motion for each box:
For box 1:
Sum of forces in the vertical direction = 0
N - W1 = 0
N = W1
N = 55 N
So, the normal force that the table exerts on box 1 is 55 N.
For box 1:
- The weight of box 1 is acting downward with magnitude W1 = 55 N.
- The normal force exerted by the table on box 1 is acting upward and is what we need to find.
For box 2:
- The weight of box 2 is acting downward with magnitude W2 = 35 N.
- The tension T in the rope is acting upward.
For box 3:
- The weight of box 3 is acting downward with magnitude W3 = 28 N.
- The tension T in the rope is acting downward.
Since the rope is massless and frictionless, the tension T is the same for boxes 2 and 3.
Now, we can set up equations of motion for each box:
For box 1:
Sum of forces in the vertical direction = 0
N - W1 = 0
N = W1
N = 55 N
So, the normal force that the table exerts on box 1 is 55 N.