A2.0 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?

tension is 37.8 N

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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

e/c, where e is its total energy