Question

A 5-lbs force is applied to the handles of the vise grip. Determine the magnitude of the compressive force developed on the smooth bolt shank A at the jaws.

Answer 1

Concepts and reason

The external force and couple moment acting on a body can be reduced to an equivalent resultant force and resultant couple moment. When this resultant force and resultant couple moment is both equal to zero then the body is said to be in equilibrium.

The major assumption for applying these equilibrium equations is that the body remains rigid.

To apply these equilibrium equations we need to know the known and unknown forces that act on the body. When all the supports are removed by replacing them with forces that prevents the translation of body in a given direction that diagram is called free body diagram.

Fundamentals

Write the equilibrium equations.

$\begin{array}{l}\\{F_R} = \sum {\bf{F}} = 0\\\\{\left( {{M_R}} \right)_O} = \sum {{{\bf{M}}_O}} = 0\\\end{array}$

Here, the resultant force is ${F_R}$ and the resultant moment about any arbitrary point is ${\left( {{M_R}} \right)_O}$ .

Calculate the magnitude of force using the trigonometric relation:

$\left| F \right| = \sqrt {{F_x}^2 + {F_y}^2}$

Here, the component of force in x-direction is ${F_x}$ and the component of force in y-direction is ${F_y}$ .

Sign Convention for force: Upward and right forces are positive.

Sign convention for moment: Anti clockwise moment is positive and clockwise moment is negative.

Draw the free body diagram of the bottom handle.

Calculate the angle made by the member CD along horizontal direction.

Apply the moment equilibrium condition.

Take moments about the point E.

$\begin{array}{l}\\\sum {{M_E}} = 0\\\\\left( {5\,{\rm{lb}}} \right)\left( {3\,{\rm{in}} + {\rm{1}}\,{\rm{in}}} \right) - \left( {{F_{CD}}\sin \theta } \right)\left( {1\,{\rm{in}}} \right) = 0\\\\\left( {5\,{\rm{lb}}} \right)\left( {3\,{\rm{in}} + {\rm{1}}\,{\rm{in}}} \right) - \left( {{F_{CD}}\sin 30.26^\circ } \right)\left( {1\,{\rm{in}}} \right) = 0\\\\{F_{CD}} = 39.693\,{\rm{lb}}\\\end{array}$

Take the equilibrium of the forces along the x-axis.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{E_x} - {F_{CD}}\cos \theta = 0\\\\{E_x} - \left( {39.693\,{\rm{lb}}} \right)\cos 30.26^\circ = 0\\\\{E_x} = 34.286\,{\rm{lb}}\\\end{array}$

Draw the free body diagram of the bottom jaw.

Apply the moment equilibrium condition.

Take moments about the point B.

$\begin{array}{l}\\\sum {{M_B}} = 0\\\\{N_A}\sin 20^\circ \left( {0.75\,{\rm{in}}} \right) + {N_A}\cos 20^\circ \left( {1.5\,{\rm{in}}} \right) - {E_x}\left( {1.75\,{\rm{in}}} \right) = 0\\\\{N_A}\sin 20^\circ \left( {0.75\,{\rm{in}}} \right) + {N_A}\cos 20^\circ \left( {1.5\,{\rm{in}}} \right) - 34.286\left( {1.75\,{\rm{in}}} \right) = 0\\\\{N_A} = 36\,{\rm{lb}}\\\end{array}$

Ans:

Therefore, the compressive force developed on the smooth bolt shank A at the jaws is $36\,{\rm{lb}}$ .