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 W1 = 52.4 N, W2 = 38.4 N, and W3 = 28.2 N. Determine the magnitude of the normal force that the table exerts on box 1.
To determine the magnitude of the normal force that the table exerts on box 1, we need to consider the forces acting on box 1.
1. Weight (W1): The weight force acts vertically downwards and has a magnitude of 52.4 N.
2. Normal Force (N1): The normal force is the force exerted by the table on box 1 and acts perpendicular to the table. Since the box is at rest, the normal force is equal in magnitude and opposite in direction to the weight force. Therefore, N1 = -W1 = -52.4 N.
Since the normal force and weight force are equal in magnitude but opposite in direction, the magnitude of the normal force that the table exerts on box 1 is 52.4 N.
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.
Let's start by drawing a free-body diagram for box 1:
1. Identify the forces acting on box 1:
- Weight of box 1 (W1) acting downwards (52.4 N).
- Normal force (N1) exerted by the table acting upwards.
- Tension in the rope between box 1 and box 2 (T1) acting towards the right.
- Tension in the rope between box 1 and box 3 (T2) acting upwards.
2. Apply Newton's second law in the vertical direction (up and down):
∑Fy = ma = 0 (since the box is not accelerating vertically)
The sum of the forces in the y-direction is given by:
N1 - W1 = 0
Solving for N1, we find:
N1 = W1
Therefore, the magnitude of the normal force that the table exerts on box 1 is equal to the weight of box 1, which is 52.4 N.
HUH?
If the rope is horizontal from box 2 to the pulley then the table is just carrying the weight of boxes 1 and 2
F = 52.4+38.4