Solving for Tension in a Suspended 28kg Sign

  • Thread starter Thread starter chrisfnet
  • Start date Start date
  • Tags Tags
    Sign Tension
AI Thread Summary
A 28kg sign is suspended by two massless cables, and the tension in each cable needs to be calculated. The equations derived include T1∙cos 57 + T2∙cos 36 = 274.4N and T1∙sin 57 + T2∙sin 36 = 0N. After solving, T2 is found to be 143.05N, leading to T1 being calculated as 291.33N. The user expresses confusion over the algebra involved but confirms they followed the correct substitution method to find both tensions. Verification of the solutions is requested to ensure accuracy.
chrisfnet
Messages
38
Reaction score
0

Homework Statement



Attached is a diagram of the problem.

A 28kg sign is suspended by two massless cables. Find the tension in each.

Homework Equations


The Attempt at a Solution



I'll call the cable at 57 degrees T1 and the other cable T2.

T1∙cos 57 + T2∙cos 36 = 9.8m/s2∙28kg
T1∙cos 57 + T2∙cos 36 = 274.4N

T1∙sin 57 + T2∙sin 36 = 0N

This is where I'm getting confused - rusty on my algebra with simultaneous equations. Am I on track so far though, then just solve for T1 and T2?
 

Attachments

  • Physics-Problem-Two.jpg
    Physics-Problem-Two.jpg
    4.2 KB · Views: 431
Physics news on Phys.org
T1 = (-T2∙sin 36)/sin 57

Substitute that into the first equation...

(-T2∙sin 36∙cos 57)/sin 57 + T2∙cos 36 = 274.4N
T2 = 143.05N

Then solve for T1.

T1∙cos 57 + (143.05N)∙cos 36 = 274.4N
T1 = 291.33N
 
From third equation you can write
T1 = - T2*sin36/sin57.
Substitute this value in eq. 2 and solve for T2.
 
rl.bhat said:
From third equation you can write
T1 = - T2*sin36/sin57.
Substitute this value in eq. 2 and solve for T2.

Thanks - that's exactly what I ended up doing. Not sure why I was over-complicating it. Could anyone confirm my solutions are correct?

Thanks!
 
Can anyone verify that my work is correct? Thanks!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top