Solve for the tension in a rope pulling a box

AI Thread Summary
The discussion focuses on calculating the tension in a rope pulling a 90-kg box at a 35° angle, with static and kinetic friction coefficients of 0.41 and 0.36, respectively. The initial calculation yielded a tension of 619.2 N, which is incorrect; the expected answer is 343 N. A key error identified was the incorrect sign change in the denominator of the tension formula, where adding sine instead of subtracting it leads to the correct result. The correct formula should maintain the addition of the sine term, which resolves the discrepancy. The final tension calculation is crucial for understanding the forces at play in this scenario.
y90x
Messages
47
Reaction score
0

Homework Statement



A worker is going to attempt to pull a 90-kg box across a level floor by a rope that makes an angle of 35° with the floor. The tension in the rope is gradually increased until the box just starts to move. If the coefficient of kinetic friction is 0.36 and the coefficient of static friction is 0.41 How much tension is in the rope when the block just starts to move?

Homework Equations


I’m guessing F=ma

The Attempt at a Solution


This is my work , however I thought I was doing it right but when I input in calculator I get the wrong answer. What am I doing wrong ?

U= mew (I believe that’s how it’s spelled)

Us(mg-Tsin(pheta))=Tcos(pheta)
Us(mg) - Us(Tsin(pheta))=Tcos(pheta)
Us(mg)=Tcos(pheta) + UsTsin(pheta)
T=(Us(mg))/(cos(pheta)-Ussin(pheta))
T=(0.41•90•9.8)/(cos35-0.41sin35)
T=619.2 N

The answer should be 343 Nhttps://www.physicsforums.com/attachments/215558
 
Physics news on Phys.org
y90x said:
Us(mg)=Tcos(pheta) + UsTsin(pheta)
T=(Us(mg))/(cos(pheta)-Ussin(pheta))<----------
T=(0.41•90•9.8)/(cos35-0.41sin35)
T=619.2 N

The answer should be 343 Nhttps://www.physicsforums.com/attachments/215558
why you switch the sign from + to - in the denominator? if you proceed with .../cos35+0.41sin35... you get the correct answer.
 
  • Like
Likes scottdave
Delta² said:
why you switch the sign from + to - in the denominator? if you proceed with .../cos35+0.41sin35... you get the correct answer.

I didn’t catch that , thanks !
 
y90x said:

Homework Statement



A worker is going to attempt to pull a 90-kg box across a level floor by a rope that makes an angle of 35° with the floor. The tension in the rope is gradually increased until the box just starts to move. If the coefficient of kinetic friction is 0.36 and the coefficient of static friction is 0.41 How much tension is in the rope when the block just starts to move?

Homework Equations


I’m guessing F=ma

The Attempt at a Solution


This is my work , however I thought I was doing it right but when I input in calculator I get the wrong answer. What am I doing wrong ?

U= mew (I believe that’s how it’s spelled)

Us(mg-Tsin(pheta))=Tcos(pheta)
Us(mg) - Us(Tsin(pheta))=Tcos(pheta)
Us(mg)=Tcos(pheta) + UsTsin(pheta)
***** T=(Us(mg))/(cos(pheta)-Ussin(pheta)) *******
T=(0.41•90•9.8)/(cos35-0.41sin35)
T=619.2 N

The answer should be 343 Nhttps://www.physicsforums.com/attachments/215558
Your attachment did not allow permission to view.

Now look at the step which I marked with stars. You went from adding Sine, to subtracting Sine. Try fixing that.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top