91Ó°ÊÓ

The coefficient of friction between the floor and the beams is\({\mu _{\rm{s}}} = 0.50\).

The angle made with the floor is\(\theta \).

Let the length of each beam be l.

A free body diagram (FBD) is a simplified representation of a rigid body to visualize all the force vectors acting on it.

With the help of free body diagrams, one can accurately define the object of interest to which a person is applying mechanical equations.

The following sketch depicts an FBD of one of the uniform beams.

The forces on the beam are the following:

• Weight of the beam,\(W = mg\)perpendicular to the floor
• Normal, N perpendicular to the floor
• Static friction between the floor and the beam,\({F_{{\rm{fr}}}}\)
• Component of the length of the beam in the upward direction,\(l\sin \theta \)
• Component of the length of the beam along the floor, \(l\cos \theta \)

According to Newton’s third law of motion, if the beam on the right pushes down the beam on the left, then the left beam pushes up on the right. It is given that the geometry of both beams is symmetric so that the beam contact force will be horizontal.

The normal force from the floor pushing up is equal to the force of gravity (weight) pulling down at the center of mass.

\({F_{\rm{N}}} = mg\) … (i)

The maximum frictional force on the floor acts towards the left and is given by the following:

\(\begin{array}{c}{F_{{\rm{fr}}}} = {\mu _{\rm{s}}}{F_{\rm{N}}}\\ = {\mu _{\rm{s}}}mg\end{array}\) … (ii)

Suppose the torque acts on the top of the beam suchthat\({F_{{\rm{beam}}}}\)exerts no torque\(\left( {r = 0} \right)\)and the net torque is zero. Consider the clockwise torques to be positive.

Thus, using the formula for the torque gives the following:

\(\begin{array}{c}\sum {\tau = 0} \\0 = {F_{\rm{N}}}l\cos \theta - mg\left( {\frac{l}{2}} \right)\cos \theta - {F_{fr}}l\sin \theta \end{array}\) … (iii)

Substitute equations (i) and (ii) in (iii).

\(\begin{array}{c}0 = \left( {mg} \right)l\cos \theta - mg\left( {\frac{l}{2}} \right)\cos \theta - \left( {{\mu _{\rm{s}}}mg} \right)l\sin \theta \\0 = \frac{{\cos \theta }}{2} - {\mu _{\rm{s}}}\sin \theta \\\frac{{\sin \theta }}{{\cos \theta }} = \frac{1}{{2{\mu _{\rm{s}}}}}\\\tan \theta = \frac{1}{{2{\mu _{\rm{s}}}}}\end{array}\)

\(\begin{array}{c}\theta = {\tan ^{ - 1}}\left( {\frac{1}{{2{\mu _{\rm{s}}}}}} \right)\\ = {\tan ^{ - 1}}\left( {\frac{1}{{2\left( {0.5} \right)}}} \right)\\ = {\tan ^{ - 1}}\left( 1 \right)\\ = {45^{\rm{o}}}\end{array}\)