Tension in a string (inclined plane, pulley, friction)

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  • #1
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Homework Statement



two blocks of mass m1 50kg and m2 100kg are connected by a string of negligible mass. The pulley is massless and frictionless. The coefficient of kinetic friction between the 50kg block and the incline is 0.250

the 100kg mass is the hanging mass.
theta is 37.0 deg

Homework Equations



F=ma

The Attempt at a Solution



basically we did this stuff near the beginning of the year and i can't remember fully how to do the problem. I know i have to make FBDs and i have and i also know i have use F=ma but im doing it wrong. If someone could please give me some clues i would be grateful :)
 

Answers and Replies

  • #2
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well, what have you done so far?
 
  • #3
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i was trying to avoid writing it out, but if you want to see it i will

i know what i've done actually, i was looking at it as if they were in equilibrium, so this is what i had:

T-m2g = Tsin37 + m1gsin37 - m1g - um1gsin37

so T = (m1gsin37 - m1g - um1gsin37 + m2g)/sin37

so i think i know what i did wrong anyways. But i still dont know how to do it the right way.
 
  • #4
326
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since it says "The coefficient of kinetic friction between the 50kg block and the incline is 0.250" it would indicate that they aren't in equilibrium or it would have been useless.
 
  • #5
114
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i was trying to avoid writing it out, but if you want to see it i will

i know what i've done actually, i was looking at it as if they were in equilibrium, so this is what i had:

T-m2g = Tsin37 + m1gsin37 - m1g - um1gsin37

so T = (m1gsin37 - m1g - um1gsin37 + m2g)/sin37

so i think i know what i did wrong anyways. But i still dont know how to do it the right way.

Your sings are wrong.
For the LHS, you have the upwards tension + and the downwards weight -.
So the for the RHS:
* the normal force*sin is upwards, so +
* tension*sin is downwards, so -
* mg is downwards, so -
* frictional force*sin is up, so +

(At least that's if I have the same diagram as you, you might want to attach the diagram if you have it)

Also in the following step you had two terms in T (T and Tsin coming from RHS), but you only divided by sin37° to isolate T, while it should have been (1-sin37°) [although the minus is wrong as stated above].


R.
 
  • #6
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i realize that now, im also doing a statics course where we assume everything is in equilibrium so thats what i have been doing for most of the semester. Basically i forget how to do these problems.
 
  • #7
326
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I think
T-mg=ma for the 100kg block
and that is the same tension to the 50kg block but you should consider friction and only the component of the weight parallel to the the incline.
 

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