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 = 0.149 kg located at 1 = 24.5° and a second mass m2 = 0.215 kg located at 2 = 279°. Calculate the mass m3, and location (in degrees), 3, which will balance the system and the ring will remain stationary.

Explanation / Answer

let tension in string connected to mass m1 is T1.

then T1=m1*g

let tension in string connected to mass m2 is T2.

then T2=m2*g

let tension in string connected to mass m3 is T3.

then T3=m3*g

let the angles are measured with respect to the horizontal axis (which we will term as x axis) and the vertical axis will be termed as y axis.

T1 is acting an angle 24.5 degrees with x axis.

T2 is acting at angle of 279 degrees with x axis.

T3 is acting at angle of theta degrees with x axis.

balancing forces in x axis:

T1*cos(24.5)+T2*cos(279)+T3*cos(theta)=0

==>T3*cos(theta)=-m1*g*cos(24.5)-m2*g*cos(279)=-1.6583...(1)

balancing forces in y axis:

T1*sin(24.5)+T2*sin(279)+T3*sin(theta)=0

T3*sin(theta)=-m1*g*sin(24.5)-m2*g*sin(279)=1.4755...(2)

dividing equation 2 by equation 1:

tan(theta)=-0.8898

==>theta=138.338 degrees

using the value of theta in equation 1,T3=2.2197 N

then m3=T3/g=0.2265 kg

hence value of m3 is 0.2265 kg and theta is 138.338 degrees for the system and ring to remain stationary.

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