A uniform horizontal beam with a length of 8.00 m and a weight of 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0° with the beam (Fig. 12.10a). If a 600-N person stands 2.00 m from the wall, find the tension in the cable as well as

Question

A uniform horizontal beam with a length of 8.00 m and a weight of 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0° with the beam (Fig. 12.10a). If a 600-N person stands 2.00 m from the wall, find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam.

Accepted Answer

Conceptualize this problem by imagining that the person in Figure 12.10 moves outward on the beam. It seems reasonable that the farther he moves outward, the larger the torque that he applies about the pivot and the larger the tension in the cable must be to balance this torque. Because the system is at rest, we categorize this as a static equilibrium problem. We begin to analyze the problem by identifying all the external forces acting on the beam: the 200-N gravitational force, the force T exerted by the cable, the force R exerted by the wall at the pivot, and the 600-N force that the person exerts on the beam. These forces are all indicated in the free-body diagram for the beam shown in Figure 12.10b.
When we assign directions for forces, it is sometimes helpful to imagine what would happen if a force were suddenly removed.
For example, if the wall were to vanish suddenly, the left end of the beam would move to the left as it begins to fall. This tells us that the wall is not only holding the beam up but is also pressing outward against it. Thus, we draw the vector R as shown in Figure 12.10b. If we resolve T and R into horizontal and vertical components, as shown in Figure 12.10c, and apply the first condition for equilibrium, we obtain

(1) [latex] \sum F_{x}=R \cos heta-T \cos 53.0^{\circ}=0 [/latex].

(2) [latex] \sum F_{y}=R \sin heta+T \sin 53.0^{\circ}-600 N -200 N =0 [/latex].

where we have chosen rightward and upward as our positive directions. Because R, T, and θ are all unknown, we cannot obtain a solution from these expressions alone.
(The number of simultaneous equations must equal the number of unknowns for us to be able to solve for the unknowns.)

Now let us invoke the condition for rotational equilibrium. A convenient axis to choose for our torque equation is the one that passes through the pin connection. The feature that makes this point so convenient is that the force R and the horizontal component of T both have a moment arm of zero; hence, these forces provide no torque about this point. Recalling our counterclockwise-equals- positive convention for the sign of the torque about an axis and noting that the moment arms of the 600-N, 200-N, and [latex] T \sin 53.0^{\circ} [/latex] forces are 2.00 m, 4.00 m, and 8.00 m, respectively, we obtain

(3) [latex] \sum au=\left(T \sin 53.0^{\circ} ight)(8.00 m )-(600 N )(2.00 m ) [/latex]

[latex] -(200 N )(4.00 m )=0 [/latex].

[latex] T=313 N [/latex].

Thus, the torque equation with this axis gives us one of the unknowns directly! We now substitute this value into Equations (1) and (2) and find that

[latex] R \cos heta=188 N [/latex].

[latex] R \sin heta=550 N [/latex].

We divide the second equation by the first and, recalling the trigonometric identity [latex] \sin heta / \cos heta= an heta, ext { we obtain } [/latex]

[latex] an heta=\frac{550 N }{188 N }=2.93 [/latex].

[latex] heta=71.1^{\circ} [/latex].

This positive value indicates that our estimate of the direction of R was accurate.
Finally

[latex] R=\frac{188 N }{\cos heta}=\frac{188 N }{\cos 71.1^{\circ}}=580 N [/latex].

To finalize this problem, note that if we had selected some other axis for the torque equation, the solution might differ in the details, but the answers would be the same. For example, if we had chosen an axis through the center of gravity of the beam, the torque equation would involve both T and R. However, this equation, coupled with Equations (1) and (2), could still be solved for the unknowns. Try it!
When many forces are involved in a problem of this nature, it is convenient in your analysis to set up a table. For instance, for the example just given, we could construct the following table. Setting the sum of the terms in the last column equal to zero represents the condition of rotational equilibrium.

Torque about O (N.m)	Moment arm relative to O (m)	Force component
[latex](8.00) T \sin 53.0^{\circ}[/latex]	8.00	[latex]T \sin 53.0^{\circ}[/latex]
0	0	[latex]T \cos 53.0^{\circ}[/latex]
-(4.00)(200)	4.00	200 N
-(2.00)(600)	2.00	600 N
0	0	[latex]R \sin heta[/latex]
0	0	[latex]R \cos heta[/latex]

Question 12.3: A uniform horizontal beam with a length of 8.00 m and a weig...

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