Chapter 12, Problem 35AP

A uniform beam of mass m is inclined at an angle θ to the horizontal. Its upper end (point P) produces a 90° bend in a very rough rope tied to a wall, and its lower end rests on a rough floor (Fig. P12.35). Let μ_s represent the coefficient of static friction between beam and floor. Assume μ_s is less than the cotangent of θ. (a) Find an expression for the maximum mass M that can be suspended from the top before the beam slips. Determine (b) the magnitude of the reaction force at the floor and (c) the magnitude of the force exerted by the beam on the rope at P in terms of m, M, and μ_s.

Figure P12.35

To determine

The expression for the maximum mass that can be suspended from the top surface before the beam starts slip.

Answer to Problem 35AP

The expression for the maximum mass that can be suspended from the top surface before the beam starts slip is $\frac{1}{2}m\left(\frac{2{\mu }_{\text{s}}\sin \theta -\cos \theta }{\cos \theta -{\mu }_{\text{s}}\sin \theta }\right)$.

Explanation of Solution

The mass of the beam is $m$, inclination angle of the beam is $\theta$, the coefficient of static friction between the beam and the floor is ${\mu }_{\text{s}}$, and the ${\mu }_{\text{s}}$ is less than cotangent of $\theta$.

The following figure shows the force diagram of the beam.

Figure-(I)

Formula to calculate the frictional force acting on the base of the beam is,

    $\overrightarrow{f}={\mu }_{\text{s}}\overrightarrow{n}$

Here, $\overrightarrow{f}$ is the frictional force acting on the base of the beam, ${\mu }_{\text{s}}$ is the coefficient of static friction and $\overrightarrow{n}$ is the normal force acting at point $O$.

Formula to calculate the net torque about the point $O$ is,

    $mg\left(\frac{L}{2}\cos \theta \right)+Mg\left(L\cos \theta \right)-\overrightarrow{T}\left(L\sin \theta \right)=0$

Here, $m$ is the mass of the beam, $g$ is the acceleration due to gravity, $L$ is the length of the beam, $\theta$ is the inclination angle of the beam with horizontal, $M$ is the mass hanged at the upper end of the beam and $\overrightarrow{T}$ is the tension on the rope.

Rearrange the above equation to find $T$.

    $\begin{array}{c}mg\left(\frac{L}{2}\cos \theta \right)+Mg\left(L\cos \theta \right)-\overrightarrow{T}\left(L\sin \theta \right)=0\\ \overrightarrow{T}\left(L\sin \theta \right)=mg\left(\frac{L}{2}\cos \theta \right)+Mg\left(L\cos \theta \right)\\ \overrightarrow{T}=\frac{mg\cos \theta }{2\sin \theta }+\frac{Mg\cos \theta }{\sin \theta }\end{array}$

Formula to calculate the net vertical forces is,

    $Mg+mg-\overrightarrow{n}=0$

Rearrange the above equation to find $\overrightarrow{n}$.

    $\begin{array}{c}Mg+mg-\overrightarrow{n}=0\\ \overrightarrow{n}=Mg+mg\end{array}$

Formula to calculate the net horizontal force is,

    $\overrightarrow{T}=\overrightarrow{f}$

Substitute $\frac{mg\cos \theta }{2\sin \theta }+\frac{Mg\cos \theta }{\sin \theta }$ for $\overrightarrow{T}$, ${\mu }_{\text{s}}\overrightarrow{n}$ for $\overrightarrow{f}$ and $Mg+mg$ for $\overrightarrow{n}$ in the above equation to find $M$.

    $\begin{array}{c}\left(\frac{mg\cos \theta }{2\sin \theta }+\frac{Mg\cos \theta }{\sin \theta }\right)={\mu }_{\text{s}}\left(Mg+mg\right)\\ \frac{m\cos \theta }{2\sin \theta }+\frac{M\cos \theta }{\sin \theta }={\mu }_{\text{s}}M+m{\mu }_{\text{s}}\\ M\left(\frac{\cos \theta }{\sin \theta }-{\mu }_{\text{s}}\right)=m\left({\mu }_{\text{s}}-\frac{\cos \theta }{2\sin \theta }\right)\\ M=\frac{m\left({\mu }_{\text{s}}-\frac{\cos \theta }{2\sin \theta }\right)}{\left(\frac{\cos \theta }{\sin \theta }-{\mu }_{\text{s}}\right)}\end{array}$

Further simplify the above equation to find $M$.

    $\begin{array}{l}M=\frac{m\left({\mu }_{\text{s}}-\frac{\cos \theta }{2\sin \theta }\right)}{\left(\frac{\cos \theta }{\sin \theta }-{\mu }_{\text{s}}\right)}\\ M=\frac{m}{2}\frac{\left(2{\mu }_{\text{s}}\sin \theta -\cos \theta \right)}{\left(\cos \theta -{\mu }_{\text{s}}\sin \theta \right)}\end{array}$

Conclusion:

Therefore, the expression for the maximum mass that can be suspended from the top surface before the beam starts slip is $\frac{1}{2}m\left(\frac{2{\mu }_{\text{s}}\sin \theta -\cos \theta }{\cos \theta -{\mu }_{\text{s}}\sin \theta }\right)$.

To determine

The magnitude of the reaction force at the floor.

Answer to Problem 35AP

The magnitude of the reaction force at the floor is $\left(m+M\right)g\sqrt{1+{\mu }_{\text{s}}{}^2}$.

Explanation of Solution

The mass of the beam is $m$, inclination angle of the beam is $\theta$, the coefficient of static friction between the beam and the floor is ${\mu }_{\text{s}}$, and the ${\mu }_{\text{s}}$ is less than cotangent of $\theta$.

Formula to calculate the net reaction force acting in the floor is,

    $\overrightarrow{R}=\sqrt{{\overrightarrow{n}}^2+{\overrightarrow{f}}^2}$

Here, $\overrightarrow{R}$ is the reaction force at the floor.

Substitute $Mg+mg$ for $\overrightarrow{n}$ and ${\mu }_{\text{s}}\overrightarrow{n}$ for $\overrightarrow{f}$ in the above equation to find $\overrightarrow{R}$.

    $\begin{array}{c}\overrightarrow{R}=\sqrt{{\left(Mg+mg\right)}^2+{\left({\mu }_{\text{s}}\left(Mg+mg\right)\right)}^2}\\ =\sqrt{{\left(Mg+mg\right)}^2\left(1+{\mu }_{\text{s}}{}^2\right)}\\ =\left(Mg+mg\right)\sqrt{\left(1+{\mu }_{\text{s}}{}^2\right)}\\ =g\left(m+M\right)\sqrt{1+{\mu }_{\text{s}}{}^2}\end{array}$

Conclusion:

Therefore, the magnitude of the reaction force at the floor is $\left(m+M\right)g\sqrt{1+{\mu }_{\text{s}}{}^2}$.

To determine

The magnitude of the force exerted by the beam in the rope at $P$.

Answer to Problem 35AP

The magnitude of the force exerted by the beam in the rope at $P$ is $g\sqrt{M^2+{\mu }_{\text{s}}{}^2{\left(m+M\right)}^2}$.

Explanation of Solution

The mass of the beam is $m$, inclination angle of the beam is $\theta$, the coefficient of static friction between the beam and the floor is ${\mu }_{\text{s}}$, and the ${\mu }_{\text{s}}$ is less than cotangent of $\theta$.

Since the coefficient of static friction of the floor is less than cotangent of the angle $\theta$, thus,

    $\overrightarrow{f}\le {\mu }_{\text{s}}\overrightarrow{n}$

Substitute $\frac{mg\cos \theta }{2\sin \theta }+\frac{Mg\cos \theta }{\sin \theta }$ for $\overrightarrow{f}$ and $Mg+mg$ for $\overrightarrow{n}$ in the above equation to find ${\mu }_{\text{s}}$.

    $\begin{array}{c}\left(\frac{mg\cos \theta }{2\sin \theta }+\frac{Mg\cos \theta }{\sin \theta }\right)\le {\mu }_{\text{s}}\left(Mg+mg\right)\\ {\mu }_{\text{s}}\ge \left[\frac{M+\frac{m}{2}}{M+m}\right]\cot \theta \end{array}$

Formula to calculate the magnitude of the net force exerted by the beam in the rope at point $P$ is,

    $\overrightarrow{F}=\sqrt{T^2+{\left(Mg\right)}^2}$

Here, $\overrightarrow{F}$ is the magnitude of the force exerted by the beam in the rope.

Substitute $\frac{mg\cos \theta }{2\sin \theta }+\frac{Mg\cos \theta }{\sin \theta }$ for $\overrightarrow{T}$ in the above equation to find $\overrightarrow{F}$.

    $\begin{array}{c}\overrightarrow{F}=\sqrt{{\left(\frac{mg\cos \theta }{2\sin \theta }+\frac{Mg\cos \theta }{\sin \theta }\right)}^2+{\left(Mg\right)}^2}\\ =\sqrt{{\left(\frac{m}{2}+M\right)}^2{g}^2{\cot }^2\theta +{\left(Mg\right)}^2}\\ =g\sqrt{{\left(\frac{m}{2}+M\right)}^2{\cot }^2\theta +M^2}\\ =g\sqrt{M^2+{\left(m+M\right)}^2{\left[\left(\frac{\frac{m}{2}+M}{m+M}\right)\cot \theta \right]}^2}\end{array}$

Substitute ${\mu }_{\text{s}}$ for $\left[\frac{M+\frac{m}{2}}{M+m}\right]\cot \theta$ in the above equation to find $\overrightarrow{F}$.

    $\begin{array}{c}\overrightarrow{F}=g\sqrt{M^2+{\left(m+M\right)}^2{\left[\left(\frac{\frac{m}{2}+M}{m+M}\right)\cot \theta \right]}^2}\\ =g\sqrt{M^2+{\left(m+M\right)}^2{\left[{\mu }_{\text{s}}\right]}^2}\\ =g\sqrt{M^2{\mu }_{\text{s}}{}^2{\left(m+M\right)}^2}\end{array}$

Conclusion:

Therefore, the magnitude of the force exerted by the beam in the rope at $P$ is $g\sqrt{M^2+{\mu }_{\text{s}}{}^2{\left(m+M\right)}^2}$