Consider the wedge and the brace

Consider the wedge and the brace shown in (Figure 1).

The Correct Answer and Explanation is :

To determine the force ( P ) required to lift the brace supporting the load ( F ) using the wedge, we must analyze the system’s static equilibrium, considering frictional forces at the contact surfaces.

Free-Body Diagrams (FBDs):

1. Brace (Block): The brace experiences:

• Its weight ( F ) acting downward.
• A normal force ( N_1 ) from the wedge acting perpendicular to the contact surface.
• A frictional force ( f_1 = \mu_s N_1 ) opposing the relative motion between the brace and the wedge.

1. Wedge: The wedge is subjected to:

• The applied force ( P ) horizontally to the right.
• The normal force ( N_1 ) from the brace, acting at an angle corresponding to the wedge’s inclination.
• A frictional force ( f_1 ) opposing the motion relative to the brace.
• A normal force ( N_2 ) from the ground, acting perpendicular to the base of the wedge.
• A frictional force ( f_2 = \mu_s N_2 ) opposing the motion relative to the ground.

Equilibrium Equations:

For the brace:

• Vertical forces: [ N_1 \cos(\theta) – f_1 \sin(\theta) = F ]

For the wedge:

• Horizontal forces: [ P = f_1 \cos(\theta) + N_1 \sin(\theta) + f_2 ]
• Vertical forces: [ N_2 = N_1 \cos(\theta) – f_1 \sin(\theta) ]

Solving for ( P ):

1. Express ( f_1 ) and ( f_2 ) in terms of ( N_1 ) and ( N_2 ):

• ( f_1 = \mu_s N_1 )
• ( f_2 = \mu_s N_2 )

1. Substitute ( f_1 ) into the brace’s vertical equilibrium equation to solve for ( N_1 ):

• [ N_1 \cos(\theta) – \mu_s N_1 \sin(\theta) = F ]
• [ N_1 (\cos(\theta) – \mu_s \sin(\theta)) = F ]
• [ N_1 = \frac{F}{\cos(\theta) – \mu_s \sin(\theta)} ]

1. Substitute ( N_1 ) into the wedge’s vertical equilibrium equation to find ( N_2 ):

• [ N_2 = N_1 (\cos(\theta) – \mu_s \sin(\theta)) ]
• [ N_2 = F ]

1. Substitute ( N_1 ), ( f_1 ), and ( f_2 ) into the wedge’s horizontal equilibrium equation to solve for ( P ):

• [ P = \mu_s N_1 \cos(\theta) + N_1 \sin(\theta) + \mu_s N_2 ]
• [ P = N_1 (\mu_s \cos(\theta) + \sin(\theta)) + \mu_s F ]
• [ P = \frac{F (\mu_s \cos(\theta) + \sin(\theta))}{\cos(\theta) – \mu_s \sin(\theta)} + \mu_s F ]
• [ P = F \left( \frac{\mu_s \cos(\theta) + \sin(\theta)}{\cos(\theta) – \mu_s \sin(\theta)} + \mu_s \right) ]

Conclusion:

The force ( P ) required to lift the brace is:

[ P = F \left( \frac{\mu_s \cos(\theta) + \sin(\theta)}{\cos(\theta) – \mu_s \sin(\theta)} + \mu_s \right) ]

This equation shows that ( P ) depends on the load ( F ), the wedge angle ( \theta ), and the coefficient of static friction ( \mu_s ). A smaller wedge angle or higher friction coefficient reduces the required force ( P ).

For a visual explanation and step-by-step solution of a similar wedge problem, you may find the following video helpful: