Question

A 2.0 kg bar with a length of 1.4 m is attached to a wall with a swivel mount. A 5.0 kg mass is suspended from the bar at a position 0.8 m from the pivot. The other end of the bar is held by a rope, which is attached vertically to the ceiling. What is the tension force in the rope?

Answer 1

tension is 37.8 N

Step-by-step explanation:

given data

mass of bar m1 = 2 kg

length of bar L = 1.4 m

suspended mass m2 = 5 kg

suspended object position length L2 = 0.8 m

to find out

tension

solution

we consider here bar is connected with hinge and

we know here system is equilibrium

so here net torque will be zero at joint

and mass 2 kg act at L1 = 1.4 /2 = 0.7 m

so torque = m1×g× ( L1 ) + m2 ×g× (L2) - T(L)

so

2 ×9.8 × ( 1.4/2) + 5×9.8 × ( 0.8) - T(1.4) = 0

T = 52.98 / 1.4

T = 37.8

so tension is 37.8 N