Question

1) A 50 kg box is hanging from two strings. String #1 pulls up and left, making an angle of 65^o with the horizontal on the left, and string #2 pulls up and to the right, making an angle of 80^o with the horizontal on the right. Find the tension in string #2 (on the right side)

2) A box is hanging from two strings. String #1 pulls up and left, making an angle of 70^o with the horizontal on the left, and string #2 pulls up and to the right with a force of 140 N, making an angle of 75^o with the horizontal on the right. Find the mass of the box.

Answer 1

1) let T1 and T2 are tensions in the left and right strings.

as the box is in equilibrium, net force acting on it must be zero.

Applly, Fnety = 0

T1*sin(65) + T2*sin(80) - m*g = 0

T1*sin(65) + T2*sin(80) = m*g

T1*sin(65) + T2*sin(80) = 50*9.8

T1*sin(65) + T2*sin(80) = 490 --------------(1)

now apply, Fnetx = 0

T2*cos(80) - T1*cos(65) = 0 --------(2)

on solving the above two equations we get

T1 = 148 N
T2 = 361 N <<<<<<<<<<--------------------Answer

2) Apply Fnetx = 0

T2*cos(75) - T1*cos(70) = 0

T1 = T2*cos(75)/cos(70)

= 140*cos(75)/cos(70)

= 106 N

m = (T1*sin(70) + T2*sin(75))/g

= (106*sin(70) + 140*sin(75))/9.8

= 24 kg <<<<<<<<<<--------------------Answer