Problem A uniform horizontal beam 5.00 m long and weighting 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 horizontal (Fig. 8.12a). If a person weighing 6.00 × 102 N stands 1.50 m from the wall, find the magnitude of the tension T in the cable and the force R exerted by the wall on the beam. R 53.0 300 N 600 N (b) R, T sin 53.0° R T cos 53.0° 53.0% 4.50 m→ 300 N 2.50 m- 600 N 5.00 m (a) (c)
Problem A uniform horizontal beam 5.00 m long and weighting 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 horizontal (Fig. 8.12a). If a person weighing 6.00 × 102 N stands 1.50 m from the wall, find the magnitude of the tension T in the cable and the force R exerted by the wall on the beam. R 53.0 300 N 600 N (b) R, T sin 53.0° R T cos 53.0° 53.0% 4.50 m→ 300 N 2.50 m- 600 N 5.00 m (a) (c)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Solution**
From Figure 8.12, the forces causing torques are the wall force \(\vec{R}\), the gravity forces on the beam and the man, \(w_B\) and \(w_M\), and the tension force \(\vec{T}\). Apply the condition of rotational equilibrium.
Compute torques around the pin at \(O\), so \(\tau_R = 0\) (zero moment arm). The torque due to the beam’s weight acts at the beam’s center of gravity.
Substitute \(L = 5.00 \, \text{m}\) and the weights, solving for \(T\).
\[
\sum \tau_i = \tau_R + \tau_B + \tau_M + \tau_T = 0
\]
\[
\sum \tau_i = 0 - w_B(L/2) - w_M(1.50 \, \text{m}) + TL \sin(53.0^\circ) = 0
\]
\[
-(3.00 \times 10^2 \, \text{N})(2.5 \, \text{m}) - (6.00 \times 10^2 \, \text{N})(1.50 \, \text{m}) + (T \sin 53.0^\circ)(5.00 \, \text{m}) = 0
\]
\[ T = \underline{\hspace{2cm}} \, \text{N} \]
Now apply the first condition of equilibrium to the beam.
Substituting the value of \(T\) found in the previous step and the weights, obtain the components of \(\vec{R}\).
\[
\sum F_X = R_X + T \cos 53.0^\circ = 0 \quad (2)
\]
\[
\sum F_Y = R_Y - w_B - w_M + T \sin 53.0^\circ = 0 \quad (2)
\]
\[ R_X = \underline{\hspace{2cm}} \, \text{N} \]
\[ R_Y = \underline{\hspace{2cm}} \, \text{N} \]
---
**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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a74cc31-62f4-4982-89c3-ce8d7d2cfbb2%2F1530af70-f11e-406a-b23e-cae973905cbc%2Fgs0lzir_processed.png&w=3840&q=75)
Transcribed Image Text:**Solution**
From Figure 8.12, the forces causing torques are the wall force \(\vec{R}\), the gravity forces on the beam and the man, \(w_B\) and \(w_M\), and the tension force \(\vec{T}\). Apply the condition of rotational equilibrium.
Compute torques around the pin at \(O\), so \(\tau_R = 0\) (zero moment arm). The torque due to the beam’s weight acts at the beam’s center of gravity.
Substitute \(L = 5.00 \, \text{m}\) and the weights, solving for \(T\).
\[
\sum \tau_i = \tau_R + \tau_B + \tau_M + \tau_T = 0
\]
\[
\sum \tau_i = 0 - w_B(L/2) - w_M(1.50 \, \text{m}) + TL \sin(53.0^\circ) = 0
\]
\[
-(3.00 \times 10^2 \, \text{N})(2.5 \, \text{m}) - (6.00 \times 10^2 \, \text{N})(1.50 \, \text{m}) + (T \sin 53.0^\circ)(5.00 \, \text{m}) = 0
\]
\[ T = \underline{\hspace{2cm}} \, \text{N} \]
Now apply the first condition of equilibrium to the beam.
Substituting the value of \(T\) found in the previous step and the weights, obtain the components of \(\vec{R}\).
\[
\sum F_X = R_X + T \cos 53.0^\circ = 0 \quad (2)
\]
\[
\sum F_Y = R_Y - w_B - w_M + T \sin 53.0^\circ = 0 \quad (2)
\]
\[ R_X = \underline{\hspace{2cm}} \, \text{N} \]
\[ R_Y = \underline{\hspace{2cm}} \, \text{N} \]
---
**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
Transcribed Image Text:**Example 8.7 Walking a Horizontal Beam**
**Goal**
Solve an equilibrium problem with nonperpendicular torques.
**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 horizontal (Fig. 8.12a). If a person weighing 6.00 × 10² N stands 1.50 m from the wall, find the magnitude of the tension \(\vec{T}\) in the cable and the force \(\vec{R}\) exerted by the wall on the beam.
**Figure 8.12**
- **(a)** A uniform beam attached to a wall and supported by a cable.
- **(b)** A free-body diagram for the beam.
- **(c)** The component of the free-body diagram.
**Diagrams Explanation**
- **(a)** Illustrates a side view of the setup, showing the beam attached to the wall with a cable making a 53.0° angle. A person stands on the beam, 1.50 m away from the wall.
- **(b)** Displays the forces acting on the beam: the person's weight (600 N), the beam's weight (300 N), and the tension (\(\vec{T}\)) and reaction force (\(\vec{R}\)) vectors.
- **(c)** Breaks down the forces into components based on the angle: \(T \sin 53.0°\) and \(T \cos 53.0°\). Reaction forces from the wall are labeled as \(R_x\) and \(R_y\).
**Strategy**
The second condition of equilibrium, \(\Sigma \tau_i = 0\), with torques computed around the pin, can be solved for the tension \(T\) in the cable. The first condition of equilibrium, \(\Sigma \vec{F}_i = 0\), gives two equations and two unknowns for the two components of the force exerted by the wall, \(R_x\) and \(R_y\).
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