A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the

way from the top of the boom. The boom is pivoted at the bottom, and a 2250 kg mass hangs from its top. The

angle between the ground and the boom is 55°. Find the tension in the cable and the components of the

Normal force on the boom by the floor.

Tension = 21,900N

Components of Normal force

Fnx = 17900N

Fny = 22700N

FN = 28900N

Explanation:

Tension in the cable is calculated by:

Etorque = - FBcostheta (1/2L) + FT (3/4L) - FWcostheta (L) = I&=0 static equilibrium

FTorque (3/4L) = FBcostheta (1/2L) + FWcostheta (L)

Ftorque = (Fcostheta (1/2L) + FWcosL) / (3/4L)

Ftorque = 2/3FBcostheta + 4/3FWcostheta

Ftorque=2/3 (1350) (9.81) cos55° + 2/3 (2250) (9.81) cos 55°

Ftorque = 21900N

b) components of Normal force

Efx=FNx-FTcos (90-theta) = 0 static equilibrium

Fnx=21900cos (90-55) = 17900N

Fy=FNy + FTsin (90-theta) - FB-FW=0

FNy = - FTsin (90-55) + FB+FW

FNy = - 21900sin (35) + (1350+2250) * 9.81=22700N

The Normal force

FN=sqrt (17900^2+22700^2)

FN = 28.900N