Answered: GOAL Apply the two conditions of…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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GOAL Apply the two conditions of equilibrium.

(a) A uniform beam attached to a wall and supported by a cable. (b) A force diagram for the beam. (c) The component form of the force diagram.

Three figures (labeled a, b, and c) are shown for the person-beam system.

In Figure (a), an illustration of the scenario is shown.
- The left end of a horizontal beam of length 5.00 m is connected to a pin on a vertical wall.
- A cable connects the right end of the beam to the top of the wall, creating an acute angle 53.0° with the horizontal.
- A person stands on the beam between the left end and the center.
In Figure (b), a horizontal beam is under the influence of four forces.
- A force vector R acts up and right on the left end of the beam.
- A force of magnitude 600 N acts downwards on the beam between the left end and the center.
- A force of magnitude 300 N acts downwards on the center of the beam.
- A force vector T acts up and left on the right end of the beam at an acute angle 53.0° from the horizontal. The magnitude of vector T is less than the magnitude of vector R.
In Figure (c), a horizontal beam is under the influence of several forces.
- A force of magnitude R_y acts vertically upwards on the left end of the beam.
- A force of magnitude R_x acts horizontally rightwards on the left end of the beam.
- A force of magnitude 600 N acts downwards on the beam 1.50 m from the left end of the beam.
- A force of magnitude 300 N acts downwards on the beam 2.50 m from the left end of the beam.
- A force of magnitude T sin(53.0°) acts vertically upwards on the right end of the beam.
- A force of magnitude T cos(53.0°) acts horizontally leftwards on the right end of the beam.

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 (Figure (a)). If a person weighing 6.00 10² N stands 1.50 m from the wall, find the magnitude of the tension in the cable and the force exerted by the wall on the beam.

STRATEGY See Figures (a)–(c) (Steps 1 and 2). The second condition of equilibrium, Σ?_i = 0, with torques computed around the pin, can be solved for the tension T in the cable. The first condition of equilibrium, Σ_i = 0, gives two equations and two unknowns for the two components of the force exerted by the wall, R_x and R_y.

SOLUTION

From the figure, the forces causing torques are the wall force , the gravity forces on the beam and the man, w_B and w_M, and the tension force . Apply the condition of rotational equilibrium (Step 3).

Σ?_i = ?_R + ?_B + ?_M + ?_T = 0

Compute torques around the pin at O, so ?_R = 0 (zero moment arm). The torque due to the beam's weight acts at the beam's center of gravity.

Σ?_i = 0 − w_B(L/2) − w_M(1.50 m) + TL sin (53°) = 0

Substitute L = 5.00 m and the weights, solving for T.

		−(3.00 ✕ 10² N)(2.50 m) − (6.00 ✕ 10² N)(1.50 m)
		+ (T sin 53.0°)(5.00 m) = 0

T = 413 N

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

(1)

ΣF_x = R_x − T cos 53.0° = 0

(2)

ΣF_y = R_y − w_B − w_M + T sin 53.0° = 0

Substituting the value of T found in the previous step and the weights, obtain the components of (Step 5).

R_x = 249 N R_y = 5.70 10² N

LEARN MORE

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 T and R_y. Together with Equations (1) and (2), however, the unknowns could still be found—a good exercise. In this example, notice the steps of the Problem-Solving Strategy could be carried out in the explicit recommended order.

QUESTION What happens to the tension in the cable if the man in Figure (a) moves farther away from the wall?

The tension would increase.The tension would decrease.

PRACTICE IT

Use the worked example above to help you solve this problem. A uniform horizontal beam 5.00 m long and weighting 3.06 10² 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 (Figure (a)). If a person weighing 6.04 10² N stands 1.40 m from the wall, find the magnitude of the tension in the cable and the force exerted by the wall on the beam.

T	= N
R_x	= N
R_y	= N

EXERCISEHINTS: GETTING STARTED | I'M STUCK!

A person with mass 55.0 kg stands d = 1.80 m away from the wall on a x = 5.55 m beam, as shown in the figure below. The mass of the beam is 40.0 kg. Find the hinge force components and the tension in the wire. (Indicate the direction with the sign of your answer.)

A person stands on a horizontal beam of length x a distance d from the left end. The left end of the beam connects to a wall and the right end of the beam connects to the left end of a wire that makes a 30° angle counterclockwise from the horizontal. The right end of the wire connects to a wall to the right of the beam.

T	= N
R_x	= N
R_y	= N

GOAL Apply the two conditions of equilibrium.
(a) A uniform beam attached to a wall and supported by a cable. (b) A force diagram for the beam. (c) The
component form of the force diagram.
53.0
300 N
53.0
-5.00 m
600 N
Tsin 53.0°
R,
Tcos 55.0
300 N
1.50 m-
600 N
2.50 m
PROBLEM A uniform horizontal beam 5.00 m long and weighing 3.00 x 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 (Figure (a)). If a person weighing 6.00 x 10 N stands 1.50 m from the wall, find
the magnitude of the tension T in the cable and the force Rexerted by the wall on the beam.
STRATEGY Se Figures (a)-(c) (Steps 1 and 2). The second condition of equilibrium, Zr, = 0, with
torques computed around the pin, can be solved for the tension T in the cable. The first condition of
equilibrium, EF = 0, gives two equations and two unknowns for the two components of the force exerted
by the wall, R and R.

PRACTICE IT
Use the worked example above to help you solve this problem. A uniform horizontal beam 5.00 m long
and weighting 3.06 x 10? 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 (Figure (a)). If a person
weighing 6.04 x 102 N stands 1.40 m from the wall, find the magnitude of the tension f in the cable and
the force R exerted by the wall on the beam.
T =
N
R =
R=
N
EXERCISE
HINTS: GETTING STARTED I I'M STUCK!
A person with mass 55.0 kg stands d = 1.80 m away from the wall on a x = 5.55 m beam, as shown in
the figure below. The mass of the beam is 40.0 kg. Find the hinge force components and the tension in
the wire. (Indicate the direction with the sign of your answer.)
30
T =
N
R =
N
R.
N

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