Chapter 9, Problem 61GP

(a)

To determine

To Draw: The free body diagram.

Explanation of Solution

Introduction:

Basically, there are three equilibrium conditions horizontal force equilibrium, vertical equilibrium and moment equilibrium condition.

There must be a downward force exerted by the slot at its left edge, exerting no torque, but balancing the vertical forces.

  

Conclusion:

Figure represents the free body diagram of a particular support reaction.

(b)

To determine

To Find: The magnitude of forces.

Answer to Problem 61GP

The magnitude of the right force is $F_{\text{Right}}=417\text{ N}$ , the magnitude of the left force is $F_{\text{Left}}=368\text{ N}$ and the magnitude of the self-weight is $W=49\text{ N}$ .

Explanation of Solution

Given:

Mass is $m=5\text{ kg}$ .

Formula used:

For right force apply torque formula, for left force apply equilibrium condition for vertical direction and for self-weight use the following formulas,

  $\begin{array}{l}\sum \tau =0\\ \sum \tau =\left(F_{\text{right}}\times d\right)-\left(m\times g\times \text{mid distance}\right)=0\end{array}$

Write the expression for Force equilibrium.

  $\begin{array}{l}\sum F_y=0\\ \sum F_y=F_{\text{right}}-m\times g-F_{\text{left}}=0\end{array}$

Here,

  $\tau$ is the torque,

  $F_y$ is the vertical force,

  $F_{\text{right}}$ is right side force,

  $F_{\text{left}}$ is left side force,

  $d$ is distance,

  $m\times g$ is the weight of shelf.

Calculation:

Write the expression for torque equilibrium.

  $\begin{array}{l}\sum \tau =\left(F_{\text{right}}\times 0.02\right)-\left(m\times g\times 0.17\right)=0\\ F_{\text{right}}=\frac{5\times 9.8\times 0.17}{0.02}=417\text{ N}\end{array}$

Write the expression for the force equilibrium.

  $\begin{array}{l}\sum F_y=F_{\text{right}}-m\times g-F_{\text{left}}=0\\ F_{\text{left}}=F_{\text{right}}-m\times g=417-(5\times 9.8)=368\text{ N}\end{array}$

Write the expression for self-weight.

  $\begin{array}{c}W=m\times g\\ W=5\times 9.8\\ =49\text{ N}\end{array}$

Conclusion:

Thus,the magnitude of theright force is $F_{\text{Right}}=417\text{ N}$ , the magnitude of the left force is $F_{\text{Left}}=368\text{ N}$ and the magnitude of the self-weight is $W=49\text{ N}$ .

(c)

To determine

To Find:The torque exerted by the support.

Answer to Problem 61GP

The torque exerted by the support about the left end of the shelf is $8.34\text{N}\text{.m}$ .

Explanation of Solution

Given:

Distance, $d=2\text{ cm = }0.02\text{ m}$ .

Force at right shelf, $F_{\text{right}}=417\text{ N}$ .

Formula used:

Applying the torque to the left end of the shelf, following formula is used

  $\tau =F_{\text{right}}\times \text{distance }(d)$

Here,

  $\tau$ is the torque,

  $F_{\text{right}}$ is right side force,

  $d$ is distance.

Calculation:

Substitute the given values in the above equation, the torque is

  $\begin{array}{c}\tau =417\times 0.02\\ =8.34\text{ N}\cdot \text{m}\end{array}$

Conclusion:

Thus, the torque exerted by the support about the left end of the shelf is $8.34\text{ N}\cdot \text{m}$ .