Chapter 7, Problem 7.8P

The brake pads at C and D are pressed against the cylinder by the spring BF. The coefficient of static friction between each pad and the cylinder is 0.15. Find the smallest tension in the spring that would prevent the cylinder from rotating when the clockwise couple $M=2200\text{lb}\cdot \text{in}\text{.}$ is applied. Neglect the weights of the members.

To determine

Smallest tension in the spring

Answer to Problem 7.8P

The smallest tension $P$ is equal to $583.28lb$ that would prevent cylinder from rotating.

Explanation of Solution

Given information:

The co-efficient of static friction between pad and cylinder is 0.15.

Couple M is equal to $2200lb.in$.

If there is no relative motion between two surfaces that are in contact, The relationship between normal force $N$ and friction force $F$ would be:

  $F={\mu }_{s}N$

  ${\mu }_s$ - Co-efficient of static friction

Steps to follow in the equilibrium analysis of a body are:

1. Draw the free body diagram.

2. Write the equilibrium equations.

3. Solve the equations for the unknowns.

Calculation:

FBD of EF

Assume $E_x,E_y$ as the reaction force components at point E.

Assume $N_2,P$ as the reaction force at point D and tension in the spring.

For the equilibrium of above section, the bending moment about point E is equal to zero.

  ${\displaystyle \sum M_E=0}$

  $\begin{array}{l}P\left(20in\right)-N_2\left(8in\right)+0.15N_2\left(4in\right)=0\\ N_2=2.703P\end{array}$

FBD of AB

Assume $A_x,A_y$ as the reaction force components at point A.

Assume $N_1,P$ as the reaction force at point C and tension in the spring.

For the equilibrium of above section, the bending moment about point A is equal to zero.

  ${\displaystyle \sum M_A=0}$

  $\begin{array}{l}P\left(20in\right)-N_1\left(8in\right)-0.15N_1\left(4in\right)=0\\ N_2=2.326P\end{array}$

FBD of the cylinder

For the equilibrium of above section, the bending moment about point O is equal to zero.

  ${\displaystyle \sum M_O=0}$

  $0.15N_2\left(5in\right)+0.15N_1\left(5in\right)-2200lb.in=0$

Substitute and solve

  $\begin{array}{l}0.75\left(2.326P+2.703P\right)-2200=0\\ P=583.28lb\\ \end{array}$

Conclusion:

The smallest tension $P$ is equal to $583.28lb$ that would prevent cylinder from rotating.