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2 pulley system with three masses. Find the two outside masses.

  • Thread starter manderzz
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Okay, so I did a lab where we had a 2 pulley system that involved a string and three masses. The string is set up so it is in the shape of an m when put on the pulleys. The known mass is put on the middle string (m3) hanging down and the two outside masses (m1) and (m2) are unknown and are not equal to each other. The middle mass is .200 kg. The angle between T1 and T2 is 80 deg the angle between T1 and T3 is 150 deg and the angle between T2 and T3 is 130 deg. I need to find M1 and M2. I drew a fbd and know that T1y and T2y must be equal to the magnitude of T3. Thank you in advance.



T1cosθ+T2cosθ+T3=0



I have been trying to work with my free body diagram to figure out the T1y and T2y values by using Pythagoras. I know that the middle mass has a weight of (9.8)(.200 kg)=1.96 N. So the T1y and T2y need to add up to equal this. I have no idea where to start, any help would be so appreciated!!
 
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  • #2
Simon Bridge
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Lets see if I have this right:

A string goes from m1, over a pulley, to m3; and another string goes from m3, over another pulley, to m2.
The whole situation is static - no motion.

Then sum the forces to zero.
m1 and m2 are hanging vertically - held by tensions T1 and T2 respectively.
Which gives you T1 and T2 right away - in terms of m1 and m2.

m3 is hanging from two angled strings, tensions T1 and T2 again - since the tension has to be the same over each pulley. I don't know how you get 3 tensions.

T1 acts upwards at angle A to the vertical and T2 acts unpwards at angle B to the vertical.
The remaining force is the weight - vertically downwards.

The horizontal and vertical components must sum to zero.
You need to use trigonometry to get the components of T1 and T2.
 
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  • #3
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Yes, that is the set up. The strings are held together by a knot if that makes a difference, and I have drawn a fbd of the known forces on the knot. Thank you, I will try it again with the information you have given.
 

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