A uniform shelf 1.8 meters long with mass 15 kg is mounted by a small hinge on a wall. The beam is held in a horizontal position by a cable that makes an angle of 37°. There is a 4.9-kg round light hanging down and is 0.55 meters from the wall. What is the tension in the supporting cable?

 T = 153.72 N

Explanation:

For this exercise we must use the conditions of translational and rotational equilibrium.

Let's set a frame of reference on the hinge, start by writing the rotational equilibrium relationship, suppose counterclockwise rotation is positive

We look for the components of the cable tension with trigonometry

           cos 37 = Tₓ / T

           sin 37 = / T

           Tₓ = T cos 37

           T_{y} = T sin 37

the expression for rotational equilibrium is

        T_{y} L + Tₓ 0 - W L / 2 - W_light  0.55 = 0

where L is the length L=  1.8 m,

        T_{y} = (W L/2 + W_lght 0.55) / L

        T sin 37 = Mg /2 + m_light g 0.55 / L

        T = (M / 2 + m_light 0.55 / L) g / sin 35

 let's calculate

        T = (15/2 + 4.9 0.55 / 1.8) 9.8 / sin 35

         T = 153.72 N

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2