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 = 0.161 kg located at 1 = 28.5° and a second mass m2 = 0.227 kg located at 2 = 295°. Calculate the mass m3, and location (in degrees), 3, which will balance the system and the ring will remain stationary.

Explanation / Answer

let the tension in the three strings be T1, T2 and T3.

as the masses are at rest,

then T1=m1*g=1.5778 N

let the table be divided along two axes, x axis is the axis along which angle is 0 degree and y axis is perpendicular to x axis.

then component of T1 along x axis=T1*cos(28.5)=1.3866 N

component of T1 along y axis=T1*sin(28.5)=0.75286 N

T2=m2*g=2.2246 N

x component=T2*cos(295)=0.94 N

y component=T2*sin(295)=-2.0162 N

let angle of T3 is theta.

then for static equilibrium, addition of all the x components will be zero as well as addition of all the y components will be zero.

then T3*cos(theta)+1.3866+0.94=0

==>T3*cos(theta)=-2.3266 N

similarly, for y component addition:

T3*sin(theta)+0.75286-2.0162=0

==>T3*sin(theta)=1.2633 N

then solving for T3 and theta, we get

T3=m3*9.8=2.6475

-->m3=0.27 kg

theta=151.5 degree

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