Question

A uniform bar of length 3.7 m and mass 4.5 kg is attached to a wall through a hinge mechanism which allows it to rotate freely. The other end of the bar is supported by a rope of length 6.7 m which is also connected to the wall as shown above. What force in N does the wall exert horizontally on the bar through the hinge? (Consider a force to the right positive and a force to the left negative.

Answer 1

Answer:

force exert horizontally  is 1 N

Explanation:

given data

bar length = 3.7 m

mass = 4.5 kg

rope length = 6.7 m

to find out

force exert horizontally

solution

we know here bar length and rope length that make angle θ

so here cos θ =  (3.7/6.7)

so equating the torque here to find force in horizontal direction is

Fx = T cos  θ   .........1

and in vertical direction

Tsinθ + N = mg    .............2

so here

we consider equilibrium condition

so

Fx = 0

T cos  θ = 0

T 3.7 / 6.7 = 0

T = 6.7/3.7

T = 1.81

so from equation 1

Fx = T cosθ

Fx = 1.81 ( 3.7/6.7)

Fx = 1 N

so force exert horizontally  is 1 N