Solve for the tension in a rope pulling a box

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Homework Help Overview

The discussion revolves around calculating the tension in a rope pulling a 90-kg box across a level floor, with the rope making an angle of 35° with the floor. The problem involves understanding the forces at play, including static and kinetic friction coefficients.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of equations related to friction and tension, with one participant expressing confusion over the signs used in the calculations. There is an exploration of the relationship between tension, friction, and the angle of the rope.

Discussion Status

Some participants have provided insights regarding the signs in the equations, suggesting that correcting the sign may lead to the expected answer. There is an ongoing examination of the calculations and assumptions made in the problem.

Contextual Notes

Participants note discrepancies in the expected answer versus the calculated result, indicating potential misunderstandings in the setup or application of the equations. The original poster's attachment was inaccessible for further clarification.

y90x
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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
 
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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.
 
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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.
 

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