A 100 kg box on the right and a 50 kg box on the lift are tied together by a horizontal cable. The 100 kg box is acted upon by an external force of 500 N to the right. The fritional force exerted by the floor on the 100 kg box is 30 N and the froctional force exerted by the floor on the 50 kg box is 20 N. In this problem, orient your axes such that the right direction is positive and the lift firection is negative. What is the net force on the 100 kg block?
Do this the same way as the tug of war problem.
To find the net force on the 100 kg box, we need to consider all the forces acting on it and their directions.
First, we have the external force of 500 N acting on the 100 kg box to the right. Since we have defined the right direction as positive, this force is already in the positive direction.
Next, we have the frictional force exerted by the floor on the 100 kg box. The frictional force always opposes the motion and acts in the opposite direction to the force applied. Therefore, the frictional force exerted by the floor on the 100 kg box is in the left direction, which is the negative direction according to our convention.
So, the net force on the 100 kg box is the vector sum of these forces. Since the force and frictional force are acting along the same line, we can simply subtract the frictional force from the force to find the net force.
Net force on the 100 kg box = Force - Frictional force
= 500 N - 30 N
= 470 N
Therefore, the net force on the 100 kg box is 470 N in the positive (right) direction.