Answered

Join IDNLearn.com and start getting the answers you've been searching for. Discover comprehensive answers to your questions from our community of knowledgeable experts.

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 :

Nuhulawal20idn

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.

To know more about magnitude of the tension,

https://brainly.com/question/25403593

We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Discover the answers you need at IDNLearn.com. Thanks for visiting, and come back soon for more valuable insights.