Question

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed 1070 N. The coefficient of static friction between the box and the floor is 0.350. (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

Answer 1

a) The angle between the cable and the horizontal is 19°

b) The weight of the sand and box = 3238.89N

Step-by-step explanation:

Let T be the tension on the cable

Let theta be the inclined angle at the horizontal

Coefficient of static friction is 0.350

Since box of sand is stationary,the net force is zero

FN + Tsintheta - mg = 0

Where FN is the magnitude of the normal force

mg= weight of the box

The normal force FN is given by:

FN= mg - Tsin theta

The horizontal component = Tcostheta-fs,max=0

T= costheta -us(mg- Tsintheta)=0

T= (usmg)/ (costheta + usintheta)

a) Tan^-1(us)= theta

Tan^-1(0.350) = 19°

b) mg= T(cos19° + usin19°)/us

mg= 1070×(cos19 + 0.350sin19°)/0.350

mg= 1070(0.9456+0.1139)/0.350

mg= 1070(1.059)/0.350

mg= 3,238.89N