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.