A 2.0 rigid rod with negligible weight is subject to forces in three different ways as shown in diagrams 1-3 below.
diagrams:
1- middle point: 2.0 N
distance on both sides: 1.0 m apart
forces below at the end of each side: 1.0 N
2- distance on both sides: 1.0 m apart
force above at end of side 1: 2.0 N
force below at the end of side 2: 2.0 N
3- distance on both sides: 1.0 m apart
forces below at the end of each side: 2.0 N
For the rod to be in equilibrium which of the diagrams above is /are correct and say why.
2. starts spinning
3. accelerates upward
To determine which diagrams represent a rod in equilibrium, we need to analyze the forces acting on the rod in each case.
Let's start with diagram 1. In this case, there is a force of 2.0 N acting at the middle point and forces of 1.0 N acting at the end of each side. To be in equilibrium, the sum of all the forces acting on the rod should be zero. However, in this case, the forces are unbalanced because the force at the middle point (2.0 N) is not counteracted by an equal and opposite force. Therefore, diagram 1 does not represent a rod in equilibrium.
Moving on to diagram 2. Here, there is a force of 2.0 N acting above at the end of side 1 and a force of 2.0 N acting below at the end of side 2. Again, we need the sum of all the forces to be zero for equilibrium. In this case, the forces are balanced because the force above (2.0 N) is counteracted by the force below (2.0 N). So, diagram 2 represents a rod in equilibrium.
Finally, let's consider diagram 3. In this case, there are forces of 2.0 N acting at the end of each side, both below. Just like before, for equilibrium, the sum of all the forces acting on the rod should be zero. In this case, the forces are unbalanced because there is no equal and opposite force to counteract the forces below. Therefore, diagram 3 does not represent a rod in equilibrium.
In summary, diagram 2 is the only one that represents a rod in equilibrium because the forces are balanced with an equal and opposite force acting at the other end.