Five boxes, each having weigh 100N, are stacked up. The bottom box is the fith from the top. What is the magnitude of the normal drove upward exerted by the fourth box from the top on the third box from the top
What is a "normal drove"?
Do you mean normal FORCE?
To find the magnitude of the normal force exerted by the fourth box from the top on the third box from the top, we need to consider the forces acting on the third box.
Let's label the boxes from top to bottom as Box 1, Box 2, Box 3, Box 4, and Box 5. We know that each box has a weight of 100N, and the bottom box is the fifth from the top. So, the weight of each box from top to bottom is:
Box 1: 100N
Box 2: 100N
Box 3: 100N
Box 4: 100N
Box 5: 100N
Since the boxes are stacked one on top of the other, the weight of each box creates a downward force on the box below it. In other words, Box 1 exerts a downward force of 100N on Box 2, Box 2 exerts a downward force of 100N on Box 3, and so on.
Now, to find the normal force exerted by the fourth box (Box 4) on the third box (Box 3), we need to consider the equilibrium of forces on Box 3. In this case, the downward force from Box 2 and the upward normal force from Box 4 should balance each other out since there is no vertical acceleration.
Since the weight of Box 3 is 100N, the downward force from Box 2 is also 100N. Therefore, to achieve equilibrium, the normal force exerted by Box 4 on Box 3 should also be 100N.
So, the magnitude of the normal force exerted by the fourth box from the top on the third box from the top is 100N.