The boom crane in the diagram is in static equilibrium and
weighs 2000 N with its center of gravity in the middle of its
length. It is supported by a frictionless pivot at the left end and
a cable connected to the center of the boom and is supporting
a load of 5000N at the end.
a) Draw a freebody diagram of the boom crane.
b) Write the three equations of static equilibrium.
c) What is the magnitude of the tension in the cable?
d) What is the magnitude of the horizontal and vertical components of the force at the pivot point?
The three equations are sum moments, horizontal forces, and vertical forces.
Those will let you find tension
a) To draw a free body diagram of the boom crane, we need to represent all forces acting on it as vectors. Here's a breakdown of the forces:
1. Weight of the boom crane (2000 N): This force acts vertically downwards from the center of gravity.
2. Tension in the cable: This force acts upwards and towards the right from the center of the boom.
3. Load force (5000 N): This force acts vertically downwards from the end of the boom.
b) The three equations of static equilibrium are as follows:
1. Sum of moments: The sum of the moments acting on an object in equilibrium must be zero. In this case, the sum of the clockwise moments must equal the sum of the counterclockwise moments. The equation for this is:
Sum of clockwise moments = Sum of counterclockwise moments
2. Horizontal forces: The sum of horizontal forces acting on the object must be zero. In other words, the forces in the horizontal direction must balance each other out. The equation for this is:
Sum of horizontal forces = 0
3. Vertical forces: The sum of vertical forces acting on the object must also be zero. The forces in the vertical direction must balance each other out. The equation for this is:
Sum of vertical forces = 0
c) To find the magnitude of the tension in the cable, we can use the equation for vertical forces. Set the sum of the vertical forces to zero and solve for the tension in the cable. Since the weight of the boom crane and the load force are both vertical forces, the equation becomes:
Tension - Weight of the boom crane - Load force = 0
Substituting the given values, we get:
Tension - 2000 N - 5000 N = 0
Simplifying the equation, we find that the tension in the cable is:
Tension = 7000 N
d) To find the magnitude of the horizontal and vertical components of the force at the pivot point, we need to separate the tension force into its horizontal and vertical components.
Since the tension force is acting at an angle, we can use trigonometry to determine the horizontal and vertical components. Let's assume the angle between the tension force and the horizontal axis is θ.
The magnitude of the horizontal component is given by:
Horizontal component = Tension * cos(θ)
The magnitude of the vertical component is given by:
Vertical component = Tension * sin(θ)
To calculate these values, we need the angle θ. Unfortunately, the diagram or additional information is needed to determine the angle accurately. Once the angle is known, the horizontal and vertical components can be calculated using the above equations.