Question

A 400kg steel beam with a length of 2.0m is horizontally suspended by two cables. On that beam is a 150kg crate placed 0.50m from the right edge. what is the value of -0-(angle between ceiling ans cable2)?

Answer 1

Let T1 and T2 are the tesions in the two cables.

As the total system is in equilibrium, net force and net torque should be equal to zero.

Net Torque about right end = 0

m1*g*0.5 + m2*g*1 - T1*sin(30)*2 = 0

T1 = 0.5*m1*g + m2*g

= 0.5*150*9.8 + 400*9.8

= 4655 N

Fnet = 0

T2*cos(phi) - T1*cos(30) = 0

T2*cos(phi) = T1*cos(30)

= 4655*cos(30)

= 4031.3 N ----(1)

T1*sin(30) + T2*sin(phi) - m1*g - m2*g = 0

0.5*T1 + T2*sin(phi) = (m1+m2)*g

T2*sin(phi) = (m1+m2)*g - 0.5*T1

= (150+400)*9.8 - 0.5*4655

= 3062.5 N --(2)

take equation(2)/equation(1)

tan(phi) = 3062.5/4031.3

= 0.76

phi = tan^-1(0.76)

= 37.2 degrees <<<<<<<<---------Answer