A 1500-N uniform boom at = 61.0° to the horizontal is supported by a cable at an angle θ = 29.0° to the horizontal as shown in the figure below. The boom is pivoted at the bottom, and an object of weight w = 2350 N hangs from its top.
To solve this problem, we need to analyze the forces acting on the boom and determine the tensions in the cable and the pivot force.
1. Draw a free body diagram of the boom:
- Draw the boom as a straight line with a length of 1500-N.
- Draw the weight force acting downward at the top of the boom with a magnitude of 2350 N.
- Draw the tension force in the cable, which pulls the boom upwards and to the left at an angle of 29.0° to the horizontal.
- Draw the pivot force, which balances the tension in the cable and the weight force, and acts in the opposite direction to the tension force.
2. Resolve the weight force:
- Break the weight force into vertical and horizontal components:
- The vertical component is given by w * cos(61.0°), where w = 2350 N.
- The horizontal component is given by w * sin(61.0°).
3. Analyze the vertical forces:
- In the vertical direction, the tension force and the vertical component of the weight force act in opposite directions.
- The tension force, T, can be calculated as T = w * cos(61.0°).
4. Analyze the horizontal forces:
- In the horizontal direction, only the horizontal component of the weight force acts.
- The pivot force must balance out the horizontal component of the weight force.
- Therefore, the pivot force, P, equals the horizontal component of the weight force, which is given by w * sin(61.0°).
5. Calculate the values:
- Using the given values:
- θ = 29.0° (angle between the cable and the horizontal)
- w = 2350 N (weight of the object)
- Calculate the tension force, T = w * cos(61.0°).
- Calculate the pivot force, P = w * sin(61.0°).
Now you have the values for the tension force, T, and the pivot force, P, which will allow you to analyze the equilibrium of the boom and solve any further questions related to it.