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A 10-meter-long, 150 kg beam extends horizontally from a wall.One end of the beam is fixed to the wall and the other end is attached to the same wall by a cable that makes an angle of 60° with the horizontal. A 75 kg sign is hung from the beam 2.50 meters from the wall.


Determine the magnitude of the tension, in [N] on the cable necessary to keep the system in equilibrium.

Sagot :

nuhulawal20

Answer:

the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

Explanation:

Given that;

Length L = 10 m

mass of beam m_b = 150 kg; weight W_beam = 150×9.8

mass of sign m = 75 kg

distance of sign hung from the beam from the wall d = 2.50 m

angle ∅ = 60°

g = 9.8 m/s²

Now,

Torque acting at one end of the beam will be;

[tex]T_{net}[/tex] = Tsin∅ × L - mg(d)-W × (L/2)

for equilibrium, [tex]T_{net}[/tex] = 0

therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)

so we substitute

Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0

Tsin(60°) × 10 - 1837.5 - 7350 = 0

Tsin(60°) × 10 - 9187.5 = 0

Tsin(60°) × 10 = 9187.5

divide both side by 10

Tsin(60°) = 918.75

T × 0.8660 = 918.75

T = 918.75 / 0.8660

T = 1060.9 N

Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

The magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.

 

Torque acting at one end of the beam,  

= Tsin∅ × L - mg(d)-W × (L/2)

When equilibrium  = 0  

Tsin∅ × L - mg(d)-W × (L/2) = 0

Where,

L - Length  =  10 m

m - mass of sign bord= 75 kg

g- gravitational accelaration = 9.8 m/s²

W - weight of beam =  150×9.8 = 1470 kg

 

Put the values in the formula,

Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0  

Tsin(60°) × 10 - 1837.5 - 7350 = 0  

Tsin(60°) × 10 - 9187.5 = 0  

Tsin(60°) × 10 = 9187.5    

Tsin(60°) = 918.75  

T × 0.8660 = 918.75  

T = 918.75 / 0.8660  

T = 1060.9 N

Therefore, the magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.

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