Q80GP When a wood shelf of mass 6.6 kg... [FREE SOLUTION]

Chapter 9: Q80GP (page 230)

^{When a wood shelf of mass 6.6 kg is fastened inside a slot in a vertical support as shown in Fig. 9–92, the support exerts a torque on the shelf. (a) Draw a free-body diagram for the shelf, assuming three vertical forces (two exerted by the support slot—explain why). Then calculate (b) the magnitudes of the three forces and (c) the torque exerted by the support (about the left end of the shelf).}

Short Answer

Expert verified

(a) The free body diagram is given in the solution given. (b) The magnitudes of the three forces are \({F_{{\rm{right}}}} = 582.1\;{\rm{N}}\), \({F_{{\rm{left}}}} = 517.42\;{\rm{N}}\), \(W = 64.6\;{\rm{N}}\), and (c) the torque exerted is \(\tau = 11.6\;{\rm{N}} \cdot {\rm{m}}\).

Step by step solution

Given data

The mass is\(m = 6.6\;{\rm{kg}}\).

The length of the shelf is\(L = 32\;{\rm{cm}}\).

The width of the shelf is \(w = 3\;{\rm{cm}}\).

The distance inside the vertical slot is \(d = 2\;{\rm{cm}}\).

Understanding net torque

In this problem, the net torque on the self is zero. While calculating the magnitudes of the three forces, consider torque about the left end of the shelf.

Free body diagram of the shelf

On the shelf, a downward force will be exerted due to its weight. There must be an upward force on the shelf and a counter-clockwise torque exerted by the slot for the shelf to be in equilibrium.

The free body diagram is as follows:

Calculation of the net torque on the shelf

The relation to find the net torque can be written as:

\(\begin{array}{c}\Sigma \tau = 0\\{F_{{\rm{right}}}}\left( {2\,{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right) - mg \times \left( {18\,{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right) = 0\end{array}\)

Here,\(g\)is the gravitational acceleration and\({F_{{\rm{right}}}}\)is the vertical force on the left side.

On plugging the values in the above relation, you get:

\(\begin{array}{c}{F_{{\rm{right}}}}\left( {2\,{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right) - \left( {{\rm{6}}{\rm{.6}}\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {18\,{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right) = 0\\{F_{{\rm{right}}}} = 582.1\;{\rm{N}}\end{array}\)

Calculation of the vertical forces

The relation to find the vertical forces can be written as:

\(\begin{array}{c}\Sigma {F_{\rm{y}}} = 0\\{F_{{\rm{right}}}} - {F_{{\rm{left}}}} - mg = 0\end{array}\)

Here,\({F_{{\rm{left}}}}\)is the upward force on the left side.

On plugging the values in the above relation, you get:

\(\begin{array}{c}\left( {582.1\;{\rm{N}}} \right) - {F_{{\rm{left}}}} - \left( {{\rm{6}}{\rm{.6}}\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) = 0\\{F_{{\rm{left}}}} = 517.42\;{\rm{N}}\end{array}\)

The relation to find the weight is given by:

\(\begin{array}{l}W = mg\\W = \left( {{\rm{6}}{\rm{.6}}\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\\W = 64.6\;{\rm{N}}\end{array}\)

Thus,\({F_{{\rm{right}}}} = 582.1\;{\rm{N}}\),\({F_{{\rm{left}}}} = 517.42\;{\rm{N}}\),and\(W = 64.6\;{\rm{N}}\)are the required forces.

Calculation of the torque exerted by the support

The relation to find the torque can be written as:

\(\tau = {F_{{\rm{right}}}} \times d\)

On plugging the values in the above relation, you get:

\(\begin{array}{l}\tau = \left( {582.1\;{\rm{N}}} \right) \times \left( {2\,{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)\\\tau = 11.6\;{\rm{N}} \cdot {\rm{m}}\end{array}\)

Thus, \(\tau = 11.6\;{\rm{N}} \cdot {\rm{m}}\) is the required torque.