please help To better understand the conccpt to static equilibrium a laboratory

Question

please help

To better understand the conccpt to static equilibrium a laboratory procedure asks the student to make a calculation before performing the experimental. The apparatus consists of a found level in the center of which is a massless ring held in place with a pin. 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 of each string. There is a mass of m1 = 154 g located at theta = 26.2degree and a second mass of m3 = 205 g located at theta = 275degree. Calculate the mass m3 and location (in degrees), theta3 which will balance the system so the pm can be removed and the ring will remain stationary. The acceleration due to gravity is g = 9.81 m/s2.

Explanation / Answer

Let Theta1=a

Theta2=b

To remove the pin all the gravitational masses have to cancel each other.

Let m3 be the mass of the third body which is located at an angle theta=c

Thus taking the horizontal component-

m1(g)[cos(a)] + m2(g)[cos(b)]+m3(g)[cos(c)]=0

g gets cancelled

m3[cos(c)]= -0.154[cos(26.2)] - 0.205[cos(275)]= -0.15604471619

m3[cos(c)]= -0.156..............(Equ1)

Taking the vertical component-

m3[sin(c)]=m1[sin(a)] + m2[sin(b)]= -0.154[sin(26.2)] - 0.205[sin(275)]=0.136

m3[sin(c)]= 0.136..............(Equ2)

Dividing equ 1 by equ 2-

cot(c)= (-0.156)/(0.136)= -1.147

c=cot inverse(-1.147)=138.9

or c=138.9-180= -41.1

But cos(c)<0 because from equ 1 m3= -0.156/cos(c) and m3 >0

Thus c can't be equal to -41.1

Thus, c=138.9 degrees

From equ 1-

m3= -0.156/cos(c)=0.207 kg

Thus, m3=207g

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