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Find the tension in the rope pulling a box at an angle

  • #1

Homework Statement



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Homework Equations




The Attempt at a Solution



Here is what i have done so far:

Fx = cos 45 T - Ff =ma
Fy=sin 45 T -mg+ Fn =ma
I found ff= uk * Fn = 0.5 *sin 45 +mg
fn = sin 45 +mg

Isolated for T in the x and y:
Tx = ma + Ff / cos 45 = 13.404
Ty = ma +mg - Fn / sin 45 = 83.938

found the resultant for T: sqrt( (13.404)^2 + (83.938)^2 ) = 85 N????


What am i doing wrong? i've been stuck on this for 3 +hours now with no help :(

the answer is meant to be E.
 
Last edited:

Answers and Replies

  • #2
andrewkirk
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Fy=sin 45 T -mg=ma
Presumably ##F_y## denotes the vertical component of net force on the box. That '=ma' should not be there. The acceleration is strictly horizontal, so ##a## shoud not appear in this equation about vertical components. Also, note that writing ##F_y## this way, in which the positive direction is up, it will be negative, because there cannot be a net upward force on the box, otherwise it would move upwards off the floor rather than horizontally along it.
I found ff= uk * Fn = 0.5 *sin 45 * mg
I don't know what 'uk' is. Assuming that 'ff' denotes frictional force, that force will be ##0.5F_y##, not what is written here.
 
  • #3
Presumably ##F_y## denotes the vertical component of net force on the box. That '=ma' should not be there. The acceleration is strictly horizontal, so ##a## shoud not appear in this equation about vertical components. Also, note that writing ##F_y## this way, in which the positive direction is up, it will be negative, because there cannot be a net upward force on the box, otherwise it would move upwards off the floor rather than horizontally along it.

I don't know what 'uk' is. Assuming that 'ff' denotes frictional force, that force will be ##0.5F_y##, not what is written here.
In the question, they stated the kinetic friction as 0.5 , which is what uk is.
and frictional force, ff, is uk * normal force (fn), therefore Ff = uk * Fn ? right?
I made the second equation as you said:

Fy=sin 45 T -mg+ Fn =0

solved for T and got -0.9998

i must be doing something wrong, im just not sure what
 
  • #4
haruspex
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solved for T and got -0.9998
We cannot tell where you went wrong if you do not post your working.
The two equations had up to that point look correct.
 
  • #5
okay i now understand that I got the normal force equation wrong it is:
Fn = mg -T sin 45

why is the tension force negative? isn't Ty in the upward positive direction?
 
  • #6
haruspex
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okay i now understand that I got the normal force equation wrong it is:
Fn = mg -T sin 45

why is the tension force negative? isn't Ty in the upward positive direction?
Yes, I understood that you had got that equation correct, but that still does not show me how you got from there to a wrong answer. Please show all your steps.
 

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