Question

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?

Short answer

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,$\mu =0.35$

Step-by-step solution

Understanding the given information

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 $1.50\text{\hspace{0.17em}m}$

Concept and formula used in the given question

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

$\begin{array}{l}\sum F_x=0\\ \sum F_y=0\\ \sum \tau =0\end{array}$

(a) Calculation for the magnitude P

For the angle between floor and beam

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

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

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

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

(b) Calculation for the magnitude of the (net) force of the floor on the beam

If we see the diagram, we can write,

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

(c) Calculation for the minimum value the coefficient of static friction between beam and floor can have for the beam not to slip at this instant

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

So,

$\begin{array}{l}{F}_{fric}=\mu F_N\\ \mu =\frac{F_{fric}}{F_N}\\ \mu =\frac{120N}{340N}\\ \mu =0.35\end{array}$