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 935 n. the coefficient of static friction between the box and the floor is 0.320. (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?

1168.7N

Explanation:

The Normal --- N -m*g + T*cos(θ) = 0...so N = m*g - T*sin(θ) ..

Therefore frictional force f = µ*(m*g - 935*sin(θ))

So the forces in the horizontal are T*cos(θ) - f = 0

935*cos(θ) = µ*(m*g - 936*sin(θ))

rearranging 935*cos(θ) + 935*µ*sin(θ) = µ*m*g

or cos(θ) + µ*sin(θ) = µ*m*g/934

to find the angle for maximum mass take the derivative of mass with respect to θ and set it to 0

So -sin(θ)*dθ + µ*cos(θ)*dθ = µ*g/935*dm

so -sin(θ) + µ*cos(θ) = µ*g/935*dm/dθ = 0

So sin(θ) = µ*cos(θ)...So tan(θ) = µ

Therefore θ = arctan(µ) = arctan(0.33) = 17.45 degrees

Now using cos(θ) + µ*sin(θ) = µ*m*g/935 we solve for m*g = W

W = 935*(cos(θ) + µ*sin(θ))/(µ)

= 935*(cos(17.45) + 0.32*sin(17.5))/(0.32) = 935 *( 0.95 + 0.32 * 0.2999 / 0.32)

= 1168.7 N

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft