0.34 Phy1340: angular momentum -- rotational equilibrium  (Page 4/6)

Solution:

We begin by determining the required value of the tension in the cable, T, to produce rotational equilibrium.

Solve for the value of T

Choose the hinge as the axis of rotation for the reasons given above.

For rotational equilibrium,

L*T*sin(30 degrees) - (L/2)*M*g = 0

Dividing both sides by L yields

T*sin(30 degrees) - M*g/2 = 0

Note that the actual length of the beam is not important for this computation.

Solving for T yields

T = (20kg*(9.8m/s^2)/2 )/sin(30 degrees), or

T = 196 newtons

Therefore, the tension in the cable is a force of 196 newtons directed from the end of the beam toward the wall at an angle of 30 degrees north of west.

Solve for the value of V

Now that we know the tension in the cable, we can compute the values for V and H.

For translational equilibrium in the vertical dimension,

V + T*sin(30 degrees) - M*g = 0, or

V = M*g - T*sin(30 degrees)

Inserting numeric values from above,

V = (20kg*9.8m/s^2) - 196newtons * sin(30 degrees), or

V = 98 newtons

Check the result for V

We can check this result by computing the sum of the torques about the center of the beam.

(L/2)*T*sin(30 degrees) - (L/2)*V = 0

Dividing through by L/2 yields

T*sin(30 degrees) - V = 0

Inserting numeric values yields

196*sin(30 degrees) - 98 = 0

and the values for T and V check.

Solve for the value of H

Knowing the value of T also makes it possible for us to compute the value of H.

For horizontal equilibrium,

H - T*cos(30 degrees) = 0, or

H = T*cos(30 degrees)

Inserting numeric values yields

H = 196newtons*cos(30 degrees), or

H = 169.7 newtons

A crane scenario

Draw the following schematic diagram of a crane on your graph board.

The crane consists of a boom attached to a horizontal surface with a pin. (The pin acts as a hinge and the boom can rotate around the pin.) Theboom extends upwards and to the right at an angle of 30 degrees relative to the surface.

A weight of 6000 newtons is hanging from the end of the boom.

A vertical cable is attached to a winch on an overhead beam and is also attached to the boom 2 meters from the lowerend.

Assumptions:

• The length of the boom = L = 8m
• The weight of the boom is negligible

Part 1

Draw a vector diagram showing the vertical forces being exerted on the boom.

Solution:

Three vertical forces are exerted on the boom:

1. A downward force at the lower end of the boom where the boom is attached to the pin. Label this force V.
2. An upward force at the point where the cable is attached to the boom. Label this force F.
3. A downward force at the upper end of the boom where the 6000 newton weight is attached. Label this force W.

Part 2

What is the downward force on the cable?

Let the lower end of the boom be the axis of rotation.

Compute the torques due to the cable and the weight. The sum of those torques must be zero for rotational stability.

2m * F*cos(30 degrees) - 8m*6000 newtons * cos(30 degrees) = 0

Divide both sides of the equation by cos(30 degrees)

2m*F = 8m*6000 newtons, or

F = (8M*6000newtons)/2m, or

F = 24000 newtons

Therefore, the force pulling down on the cable is 24000 newtons.