1. Two similar cylindrical polished bars weighing 5.00 N each lie next to one another in contact. A third similar bar is placed on the other two in the groove between them. Neglecting friction what horizontal force on each other is necessary to keep them together?
2. Many presses and other tools make use of the toggle, which is a pair of bars jointed together so as to form a very obtuse angle between them. When a 20 N force is exerted on the joint along the angle besector, the opposite ends of the bars may exert very larger forces on their restraints. If the angle at the joint is 170O , find the two horizontal forces necessary to hold the system in equilibrium. [Ans. 74N]
2.5 n
POTA
74
To solve these questions, we need to apply the laws of equilibrium, specifically Newton's second law and the conditions for equilibrium.
1. Two similar cylindrical bars weighing 5.00 N each are in contact. We have a third similar bar placed on the other two in the groove between them. We need to find the horizontal force required to keep them together, neglecting friction.
To solve this, we can start by considering the forces acting on the middle bar. The weight of the middle bar is in the downward direction, and there will be forces acting on it from the bars on either side.
Let's assume the horizontal force required to keep the bars together is F Newtons. Since the bars are in contact and we neglect friction, the force between the middle bar and each of the other bars will be equal to F.
Now, let's analyze the forces acting on the middle bar. The forces acting on it are the weight (downward) and the forces from the other bars (upward).
Since the bars are in equilibrium, the net force and net torque on the middle bar must be zero.
Considering forces in the horizontal direction:
F - F = 0
Therefore, F = 0 N
This means that no horizontal force is required to keep the bars together, and they will stay in contact without any external force.
2. In this problem, we have a toggle system made up of two bars jointed together to form a very obtuse angle of 170°. A 20 N force is exerted on the joint along the angle bisector. We need to find the two horizontal forces necessary to hold the system in equilibrium.
To solve this, let's first analyze the forces acting on the bars. We can assume that the bars are connected by a hinge joint, enabling them to rotate.
Considering the forces in the horizontal direction, we can break down the 20 N force into its components.
The force exerted on the joint can be split into two components, one in the horizontal direction and one in the vertical direction.
The horizontal component will be:
Horizontal force = 20 N * cos(170°)
This horizontal force needs to be balanced by forces in the opposite direction, exerted on both ends of the bars.
So, each bar will exert a horizontal force of:
F = (Horizontal force) / 2
Plugging in the values:
F = (20 N * cos(170°)) / 2
Evaluating this expression will give us the horizontal force required to hold the system in equilibrium, which will be approximately 74 N.
Hence, the two horizontal forces required to hold the system in equilibrium are approximately 74 N each.