Question

A bucket of water of mass 14.6 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.320 m with mass 11.1 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 11.0 m to the water. You can ignore the weight of the rope.
Required:
a. What is the tension in the rope while the bucket is falling?
b. What is the time of fall?
c. While the bucket is falling, what is the force exerted on the cylinder by the axle?

Answer 1

Answer:

a. T = 39.41 N

b. t = 1.76s

c. 150.78 N

Explanation:

Given:

Mass of bucket of water, Mb = 14.6 kg

Mass of cylinder, Mc = 11.1 kg

Diameter of cylinder, D =  0.320 m, or radius, r = D/2 = 0.16m

Displacement of the bucket from the top, that is the vertical displacement , y = 11.0 m

a. The tension in the rope while the bucket is falling is:

F = mg - T = ma

Where F= The force

m= mass

g= Acceleration due to gravity

T = tension in the rope

a = acceleration

T= m(g - a)

Then, calculating the angular acceleration of the pulley system in relation to its radial acceleration

T= 1/2Ma

Merging the two final equation so as to solve for a

M(g - a) = 1/2Ma

Make a the subject of the formula

Mg - Ma = 1/2Ma

1/2Ma + Ma = Mg

a (1/2 M + M) = Mg

Divide both side by (1/2 M + M)

a = Mg ÷ (1/2 M + M)

Inputing the given value in the formula above

g= 9.8m/s2

a = (14.6 kg) (9.8m/s2) ÷ 1/2 (11.1 kg) + 14.6 kg

a = 7.1007m/s2

Now it is easy to input the value into T= 1/2Ma

T = 1/2 (11.1 kg) (7.1007m/s2) = 39.41 N

B. Time of fall is:

Using one of the equation of motion

s = ut + 1/2 at^2

U = Initial velocity

t = time

a = acceleration

s= distance in this case displacement y

making t the subject of the formula

t = √(2s ÷ a)

u is 0 since the bucket starts from rest

so, t = √((2)(11.0 m) ÷ 7.1007m/s2)

t = 1.76s

c.  the force exerted on the cylinder by the axle = T + Mg

  = 42 N + (11.1 kg) (9.8m/s2)

= 150.78 N