WALKING A HORIZONTAL BEAM GOAL Apply the two conditions of equilibrium. PROBLEM A uniform horizontal beam 5.00 m long and weighing 3.00×10² N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0∘ with the

Question

WALKING A HORIZONTAL BEAM

GOAL Apply the two conditions of equilibrium.

PROBLEM A uniform horizontal beam [latex]5.00 \mathrm{~m}[/latex] long and weighing [latex]3.00 imes 10^{2} \mathrm{~N}[/latex] is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of [latex]53.0^{\circ}[/latex] with the horizontal (Fig. 8.18a). If a person weighing [latex]6.00 imes 10^{2} \mathrm{~N}[/latex] stands [latex]1.50 \mathrm{~m}[/latex] from the wall, find the magnitude of the tension [latex]\overrightarrow{\mathbf{T}}[/latex] in the cable and the components of the force [latex]\overrightarrow{\mathbf{R}}[/latex] exerted by the wall on the beam.

STRATEGY See Figure 8.18a-c (Steps 1 and 2). The second condition of equilibrium, [latex]\Sigma_{ au_{i}}=0[/latex], with torques computed around the pin, can be solved for the tension [latex]T[/latex] in the cable. The first condition of equilibrium, [latex]\sum \overrightarrow{\mathbf{F}}_{i}=0[/latex], gives two equations and two unknowns for the two components of the force exerted by the wall, [latex]R_{x}[/latex] and [latex]R_{y}[/latex].

Accepted Answer

From Figure [latex]8.18[/latex], the forces causing torques are the wall force [latex]\overrightarrow{\mathbf{R}}[/latex], the gravity forces on the beam and the man, [latex]w_{B}[/latex] and [latex]w_{M}[/latex], and the tension force [latex]\overrightarrow{\mathbf{T}}[/latex]. Apply the condition of rotational equilibrium (Step 3) :

[latex]\sum au_{i}= au_{R}+ au_{B}+ au_{M}+ au_{T}=0[/latex]

Compute torques around the pin at [latex]O[/latex], so [latex] au_{R}=0[/latex] (zero moment arm). The torque due to the beam's weight acts at the beam's center of gravity.

[latex]\sum au_{i}=0-w_{B}(L / 2)-w_{M}(1.50 \mathrm{~m})+T L \sin \left(53^{\circ} ight)=0[/latex]

Substitute [latex]L=5.00 \mathrm{~m}[/latex] and the weights, solving for [latex]T[/latex] :

[latex]\begin{aligned}&-\left(3.00 imes 10^{2} \mathrm{~N} ight)(2.50 \mathrm{~m})-\left(6.00 imes 10^{2} \mathrm{~N} ight)(1.50 \mathrm{~m}) \&+\left(T \sin 53.0^{\circ} ight)(5.00 \mathrm{~m})=0 \T &=413 \mathrm{~N}\end{aligned}[/latex]

Now apply the first condition of equilibrium to the beam (Step 4):

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

(2) [latex]\sum F_{y}=R_{y}-w_{B}-w_{M}+T \sin 53.0^{\circ}=0[/latex]

Substituting the value of [latex]T[/latex] found in the previous step and the weights, obtain the components of [latex]\overrightarrow{\mathbf{R}}[/latex] (Step 5):

[latex]R_{x}=249 \mathrm{~N} \quad R_{y}=5.70 imes 10^{2} \mathrm{~N}[/latex]

REMARKS Even if we selected some other axis for the torque equation, the solution would be the same. For example, if the axis were to pass through the center of gravity of the beam, the torque equation would involve both [latex]T[/latex] and [latex]R_{y}[/latex]. Together with Equations (1) and (2), however, the unknowns could still be found-a good exercise. In both Example 8.9 and Example 8.11, notice the steps of the Problem-Solving Strategy could be carried out in the explicit recommended order.

Question 8.11: WALKING A HORIZONTAL BEAM GOAL Apply the two conditions of e...

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