0.34 Phy1340: angular momentum -- rotational equilibrium  (Page 5/6)

Part 3

What is the magnitude and direction of the force on the pin at the bottom of the boom?

For vertical equilibrium, the vertical forces must sum to zero.

V + 24000 newtons - 6000 newtons = 0, or

V = 6000 newtons - 24000 newtons, or

V = -18000 newtons

Therefore, the force on the pin is 18000 newtons down.

Sliding a crate

A crate is being slid to the right on a horizontal floor by pulling on a rope tied to the right side of the crate.

Width of the crate = w = 4 m

Height of the crate = h = 3 m

Coefficient of kinetic friction = u = 0..5

Attachment point of rope = d = ?

Part 1

Draw a picture of the crate showing the forces that are being exerted on the crate.

Solution:

The crate is a rectangle with the dimensions given above. There are three forces acting on the crate.

There is a downward force at the bottom center of the crate, which is the weight of the crate. Label it F1.

There is a horizontal force pointing to the left at the bottom of the crate. This is the force of friction. Label it F2.

There is a horizontal force pointing to the right on the right side of the crate that is d units above the floor. This is the force that is pulling thecrate to the right. Label it F3.

Part 2

What is the maximum value for d that allows the crate to slide without tipping over?

Compute the sum of the horizontal forces.

F3 - F2 = 0, or

F3 - m*g*u = 0, or

F3 = m*g*u

Compute the torques about the bottom right corner.

(w/2)*F1 - d*F3 = 0, or

d*F3 = (w/2)*F1, or

d*F3 = (w/2)*m*g

Substitution yields

d*m*g*u = (w/2)*m*g

Simplification yields

d*u = (w/2), or

d = (w/2)/u

Inserting values yields

d = (4/2)/0.5, or

d = 4 m

Since the height of the crate is only 3 m, it is not possible to tip it over by pulling on a rope attached to the right side of the crate.

Another crane scenario

The boom of a crane is pinned at the intersection of a horizontal floor and a vertical wall. The boom slopes toward the upper right.

A mass hangs from the top end of the boom.

The top end of the boom is connected to the vertical wall by a cable that is stretched horizontally from the wall to the boom.

Length of cable = X = 4 m

Height of cable = Y = 3 m

Mass = M = 2000 kg

The weight of the boom is negligible.

Part 1

Draw the forces acting on the boom.

Solution:

There are four forces acting on the boom:

A downward force that is attributable to the mass hanging on the top end of the boom. Label this force F1.

A force pointing to the left from the top end of the boom due to the cable. Label this force F2.

A force pointing up at the bottom end of the boom. Label this force F3.

A force pointing to the right at the bottom of the boom. Label this force F4.

Part 2

Find the tension in the cable.

Solution:

For rotational equilibrium, the sum of the torques about the bottom of the boom must be zero.

F2*Y = F1*X, or

F2*Y = M*g*X, or

F2 = (M*g*X)*/Y

Inserting numeric values yields

F2 = (2000kg*(9.8m/s^2)*4m)/3m

F2 = 26133 newtons

Therefore, the tension in the cable is equal to 26133 newtons

Part 3

Find the force pushing against the bottom end of the boom along the length of the boom. Also find the angle between that force and the floor.