Question
Asked Nov 14, 2019

A 4-m long, 150-kg steel beam is attached to a wall with one end connected to a hinge that allows the beam to rotate up and down. The other end of the beam is held in a horizontal position with a cable that makes a 27° angle with the beam and is attached to the wall (see (Figure 1)).

a.What is the tension force that keeps this beam in static equilibrium?

b.  A mass of 75 kg is hung from the beam 3 meters away from the hinge (see (Figure 2)). Now what is the tension force that keeps this beam in static equilibrium?

c.What is the vertical component of the force that the hinge exerts on the beam?

 

27
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27

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27
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27

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Expert Answer

Step 1

(a) Applying balance of moments,

ΣΜ-0
(2 m)(150 kg)(9.8m/s2)-(4m)(T sin 270) = 0
(2 m)(150kg) (9.8m/s?
Т-
(4m)(sin 27)
=1618.98N
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ΣΜ-0 (2 m)(150 kg)(9.8m/s2)-(4m)(T sin 270) = 0 (2 m)(150kg) (9.8m/s? Т- (4m)(sin 27) =1618.98N

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Step 2

(b) Applying balance of moments about the hinge point,

ΣΜ-
(2m) (150 kg) (9.8m/s)+ (3m)(75kg)(9.8m/s)-(4m) (T sin 27°)- 0
(2m)(150kg)(3 m) (75kg)
(4 m) (sin 27
T (9.8m/s
=
= 2833 21N
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ΣΜ- (2m) (150 kg) (9.8m/s)+ (3m)(75kg)(9.8m/s)-(4m) (T sin 27°)- 0 (2m)(150kg)(3 m) (75kg) (4 m) (sin 27 T (9.8m/s = = 2833 21N

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Step 3

(c) Vertical component of the force that...

T sin 27+F(m+m2)g
F (150kg+75kg) (9.8m/s)-(2833.21 N) ( sin 27°)
=918.75N
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T sin 27+F(m+m2)g F (150kg+75kg) (9.8m/s)-(2833.21 N) ( sin 27°) =918.75N

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