To better understand the concept of static equilibrium a laboratory procedure as

Question

To better understand the concept of static equilibrium a laboratory procedure asks the student to make a calculation before performing the experiment. The apparatus consists of a round level table in the center of which is a massless ring. There are three strings tied to different spots on the ring. Each string passes parallel over the table and is individually strung over a frictionless pulley (there are three pulleys) where a mass is hung. The table is degree marked to indicate the position or angle of each string. There is a mass m1 = 157 g located at theta = 24.5 degree and a second mass m2 = 219 g located at theta 2 = 295degree Calculate the mass m3, and location (in degrees), theta 3, which will balance the system and the ring will remain stationary.

Explanation / Answer

Let us assume top of the round table is our XY plane and Tension of the strings are weights of respective masses.

So the diagram on two dimensional plane can seen as below.

T3 is not shown since direction of it not known yet

Now you can see system is in equilibrium along X axis as well as Y axis.

Applying Newton’s second law to the tensions on the strings,

T1+T2+T3 = 0      ………….vectorally

Along X axis,

T1x+T2x+T3x= 0     

m1gcos24.5 +m2gcos295 + T3x=0

0.157*9.8cos24.5 +0.219*9.8cos295 +T3x=0           => T3x= 2.31 N

Along X axis,

T1y+T2y+T3y= 0     

m1gsin24.5 +m2gsin295 + T3x=0

0.157*9.8sin24.5 +0.219*9.8sin295 +T3y=0           => T3y= -1.31 N

Angle between T3 and X axis = q3 = tan^-1(Ty/Tx) = tan^-1(-1.31/2.31) = -29.56 deg

Thus,

T3x= 2.31

m3gcosq3 = 2.31

m3*9.8*cos(-29.56) = 2.31       => m3= 0.271kg= 271g

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