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Q68P

Expert-verifiedFound in: Page 352

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A construction worker attempts to lift a uniform beam off the floor and raise it to a vertical position. The beam is ** ${\mathbf{2}}{\mathbf{.50}}{\mathbf{\text{\hspace{0.17em}m}}}$** long and weighs${\mathbf{500}}{\mathbf{\text{\hspace{0.17em}N}}}$ ** **. At a certain instant the worker holds the beam momentarily at rest with one end at distance ${\mathbf{\text{d}}}{\mathbf{}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.50}}{\mathbf{\text{\hspace{0.17em}m}}}$** ** above the floor, as shown in Fig. 12-75, by exerting a force ** **on the beam, perpendicular to the beam. (a) What is the magnitude P? (b) What is the magnitude of the (net) force of the floor on the beam? (c) What is the minimum value the coefficient of static friction between beam and floor can have in order for the beam not to slip at this instant?**

a) The magnitude of the applied force, $\text{F}=200\text{\hspace{0.17em}N}$

b) Net force exerted by the floor on beam, ${\text{F}}_{\text{N}}=340\text{\hspace{0.17em}N}$

c) The minimum value of the coefficient of static friction, $\mu =0.35$

The length of the beam is .$2.50\text{\hspace{0.17em}N}$

Weight of beam is $500\text{\hspace{0.17em}N}$ .

Given the distance from the floor is $1.50\text{\hspace{0.17em}m}$

**Using the condition for the static equilibrium, you can write the equation for torque. Solving this, you would get force. Using this value of force, you can find the coefficient of friction. The equations are given below**

**$\begin{array}{l}\sum {F}_{x}=0\\ \sum {F}_{y}=0\\ \sum \tau =0\end{array}$**

For the angle between floor and beam

As the beam is diagonal of a triangle and given height is its one side.

$\begin{array}{c}\theta =si{n}^{-1}\left(\frac{1.5}{2.5}\right)\\ \theta =37\xb0\end{array}$

As the system is in equilibrium, we can write the moment of force as

$\begin{array}{l}F\times L-\frac{L}{2}\times Mgcos\theta =0\\ F=\frac{Mgcos\theta}{2}\\ F=\frac{500\text{\hspace{0.17em}}N\times \mathrm{cos}37\xb0}{2}\\ F=199.6\text{\hspace{0.17em}N}\end{array}$

If we see the diagram, we can write,

$\begin{array}{c}{F}_{N}=Mg-Fcos\theta \\ =500\text{\hspace{0.17em}}N-\left(200\text{\hspace{0.17em}}N\right)\left(0.87\right)\\ =340\text{\hspace{0.17em}N}\end{array}$

$\begin{array}{c}{F}_{fric}=\mu {F}_{N}\\ =Fsin37\xb0\\ =120\text{\hspace{0.17em}N}\end{array}$

So,

$\begin{array}{l}{F}_{fric}=\mu {F}_{N}\\ \mu =\frac{{F}_{fric}}{{F}_{N}}\\ \mu =\frac{120\text{\hspace{0.17em}}N}{340\text{\hspace{0.17em}}N}\\ \mu =0.35\end{array}$

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