Two blocks hanging on an incline

  • #1
Two blocks are connected by a massless string and are held in position by another massless string along a frictionless incline (as shown in the figure). Let M1 = 39.0 kg, M2 = 37.0 kg, and θ = 36°. Calculate the tension in the string connecting the two blocks.

Calculate the tension between the wall and M2

I'm working on the M2 part fist:

When I was working out this problem, I got these two equations

Y= T*sin(36)-382.2-362.6=0
X= T*cos(36)

so I solved for T in x
T= cos(36)

Am I approaching this the right way. I'm still having problems calculating components on forces.
 

Answers and Replies

  • #2
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do you have the figure?
 
  • #3
M2 is closest to the wall
 
  • #4
mjsd
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get two equations with two unknowns: T1 and T2 for the two tensions
 
  • #5
So for T1 or the Tension between M2 and the Wall

X= cos(36)*T1=0
Y= T*sin(36)-382.2-362.6=0

For T2
x= cos(36)*(T1+T2)=0
y= sin(36)*(T1+T2)-382.2-362.6=0

Am I on the right track now?
 
  • #6
mjsd
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confusing notation... does "X=..." means for x-comp etc?
cos(36)*M2*g = N (normal reaction force for M2)
how come you have Tesion force multiplying cos and sin? probably easier to use the coord system where y direction is perpendicular to surface of incline plane and x is parallel to surface.

then just relate weight forces to the tensions
 
  • #7
Those are the components, X= is parallel to the x axis and Y is perpendicular to the X
 
  • #8
mjsd
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I believe it would be easier to use the set of coord sys that I have mentioned
 
  • #10
mjsd
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  • #11
oh gosh, this is embarasssing

do you mind just correcting one equation and I'll use that to fix the rest?
 
  • #12
mjsd
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as I understood it (but without the picture I am guessing of course), there are two known forces (the weight forces), and two unknowns (the tensions). you task is to put tensions force in terms of the weight forces. since weight forces always act vertically down, and with the "inclined" coord sys, you must first resolve the components of these weight forces in the directions of these "inclined" axes. I have already given you one equation before (which illustrates how you should resolve stuffs... although that equation itself has no use in this particular problem because there is no friction)

cos(36)*M2*g = N (normal reaction force for M2)
from this you should observe that cos(36)*M2*g is actually perpendicular to surface of incline plane. you just have to do the other directions and the fact that all forces sum to zero.
 

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