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Q38P

Expert-verifiedFound in: Page 349

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig 12-52, uniform beams A and B are attached to a wall with hinges and loosely bolted together (there is no torque of one on the other). Beam A has length${\mathit{L}}_{\mathbf{A}}\mathbf{=}\mathbf{2}\mathbf{.40}\mathbf{\text{\hspace{0.17em}m}}$ and mass${\mathbf{54}}{\mathbf{.0}}{\mathbf{\text{\hspace{0.17em}}}}{\mathit{k}}{\mathit{g}}$; beam B has mass${\mathbf{68}}{\mathbf{.0}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{}}{\mathit{k}}{\mathit{g}}$. The two hinge points are separated by distance${\mathit{d}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.80}}{\mathbf{}}{\mathit{m}}$.In unit-vector notation, what is the force on(a) beam A due to its hinge? (b) beam A due to the bolt? (c) beam B due to its hinge? And (d) beam B due to the bolt?**

** **

a) Force on beam A due to hinge is $-797\widehat{i}+265\widehat{j}$.

b) Force on beam A due to bolt is$797\widehat{i}+265\widehat{j}$ .

c) Force on beam B due to hinge is$797\widehat{i}+931\widehat{j}$ .

d) Force on beam B due to bolt is$-797\widehat{i}-265\widehat{j}$ .

- The length of beam A is,$L=2.40\text{\hspace{0.17em}}m$.
- Mass of beam A is,${m}_{A}=54\text{\hspace{0.17em}}kg$.
- Mass of beam B is,${m}_{B}=68\text{\hspace{0.17em}}kg$.
- The separation between beams is$d=1.80\text{\hspace{0.17em}}m$,.

**Write down the torque equation about the pivot point to find out the x and y components of forces. Using the values of forces found here, we can solve the remaining parts.**

The free body diagram can be drawn as below:

Total mass is,$M=(54+68)=122\text{\hspace{0.17em}}kg$.

The line of action of that downward force of gravity is$x=1.20\text{\hspace{0.17em}}m$ from the wall. The vertical distance between the hinges is $y=1.80\text{\hspace{0.17em}}m$.

From the equation of the torque at the bottom of the hinge, we can calculate the x component of force as,

${F}_{x}=-\frac{Mgx}{y}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{x}=-\frac{122\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\times 1.8\text{\hspace{0.17em}m}}{1.2\text{\hspace{0.17em}m}}\\ -\frac{122\times 9.8\times 1.8}{1.2}\cdot (\frac{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\times 1\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}m}}\times \frac{1\text{\hspace{0.17em}}N}{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}})\\ =-797\text{\hspace{0.17em}}N\end{array}$

The torque about bolt :

The weight of the beam would be acting downwards at the center of the beam. The torque caused by this weight would be balanced by torque caused by Fy in the upward direction.

${F}_{y}=\frac{{m}_{A}gx}{l}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{y}=\frac{54\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\times 1.2\text{\hspace{0.17em}m}}{2.4\text{\hspace{0.17em}m}}\\ {F}_{y}=\frac{54\times 9.8\times 1.2}{2.4}\cdot (\frac{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\times 1\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}m}}\times \frac{1\text{\hspace{0.17em}}N}{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}})\\ =265\text{\hspace{0.17em}}N\end{array}$

So unit vector notation, we can write the force as,

$\overrightarrow{F}={F}_{x}\widehat{i}+{F}_{y}\widehat{j}$

Substitute the values in the above expression, and we get,

$\overrightarrow{F}=-797\widehat{i}+265\widehat{j}$

Thus, the force on beam A due to the hinge is $-797\widehat{i}+265\widehat{j}$.

Equilibrium of horizontal and vertical forces on beam A readily yields.

X component of force due to bolt is and given as follows:

$\begin{array}{c}{F}_{1}=-{F}_{x}\\ =727\text{\hspace{0.17em}}N\end{array}$

Y component of force due to bolt is${F}_{2}$ and given as,

${F}_{2}={m}_{A}g-{F}_{y}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{2}=54\times 9.8-265\\ =529.2-265\\ =264.2\\ ~265\text{\hspace{0.17em}N}\end{array}$

So unit vector notation, we can write the force as,

$\overrightarrow{F}={F}_{1}\widehat{i}+{F}_{2}\widehat{j}$

Substitute the values in the above expression, and we get,

$\overrightarrow{F}=797\widehat{i}+265\widehat{j}$

Thus, the force on beam A due to the bolt is $797\widehat{i}+265\widehat{j}$.

Now consider combining the AOB system equilibrium of horizontal and vertical forces.X component of force due to hinge at B is${F}_{3}$

$\begin{array}{c}{F}_{3}=-{F}_{x}\\ =797N\end{array}$

Y component of force due to hinge at B is ${F}_{y}$.

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{4}=Mg-{F}_{y}\\ =122\times 9.8\\ =931N\end{array}$

So unit vector notation, we can write the force as,

$\overrightarrow{F}={F}_{3}\widehat{i}+{F}_{4}\widehat{j}$

Substitute the values in the above expression, and we get,

$\overrightarrow{F}=797\widehat{i}+931\widehat{j}$

Thus, the force on beam B due to hinge is $797\widehat{i}+931\widehat{j}$

** **

Force on beam B due to bolt:

X component of force due bolt at B is${F}_{5}$ and given as follows:

${F}_{5}=-797\text{\hspace{0.17em}}N$

Y component of force due to bolt at B is${F}_{6}$ and given as,

${F}_{6}=-265\text{\hspace{0.17em}}N$

So unit vector notation, we can write the force as,

$\overrightarrow{F}={F}_{x}\widehat{i}+{F}_{y}\widehat{j}$

So vector notation of force at B due to bolt at B is as follows:

$\overrightarrow{F}=-797\widehat{i}-265\widehat{j}$

Thus, the force on beam B due to the bolt $-797\widehat{i}-265\widehat{j}$ .

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