Question

A force of \( P = 100 \, {N} \) is applied at the ends of the pliers as shown in the figure. Neglecting friction, the force exerted by the upper jaw on the workpiece is ........... N (in integer). 

Hint:

When a force is applied at the ends of a lever (or pliers in this case), the force exerted by the other part of the lever can be found by balancing the moments (torques) about the pivot point.

Answer 1

Step 1: In this case, the problem involves a force applied at the ends of the pliers, and the force exerted by the upper jaw on the workpiece can be found by considering the principle of equilibrium and the leverage.
Step 2: The forces applied at the ends of the pliers create a moment (torque) about the pivot point where the workpiece is in contact. Since we are neglecting friction, we only need to consider the moments about the pivot.
Step 3: The force at the ends of the pliers creates a moment \( M \), which is equal to the force \( P \) multiplied by the distance from the pivot point:
\[ M = P \times 100 \, {mm} = 100 \times 100 = 10000 \, {N.mm} \] This moment is balanced by the force exerted by the upper jaw on the workpiece, which acts at a distance of 25 mm from the pivot point.
Step 4: Let \( F_{{upper}} \) be the force exerted by the upper jaw on the workpiece. The moment exerted by this force is:
\[ M = F_{{upper}} \times 25 \, {mm} \]
Step 5: Since the moments are balanced, we can equate the two moments:
\[ 10000 \, {N.mm} = F_{{upper}} \times 25 \, {mm} \] Solving for \( F_{{upper}} \):
\[ F_{{upper}} = \frac{10000}{25} = 400 \, {N} \]
Step 6: Therefore, the force exerted by the upper jaw on the workpiece is \( 400 \, {N} \).