A crane is holding a bloc of metal of 10 000kg. If the crane's arm weights 1000kg and is 10,0m long, what is the tension in cable a and how big is the force exerted on the pivot of the cranes arm.
the cable passes through the top of the cranes arm and then is vertical whilst holding the bloc of metal. The angle formed by the cable on the left side with the arm is 10* and is of 45* on the right side of the arm, between it and the bloc of metal + cable. I'm sorry but this is the best way for me the explain the image. The crane is pointing to the right. ANd when do I know when to use sin or cos, I never understand it.
Thank you
To find the tension in cable A and the force exerted on the pivot of the crane's arm, we can break down the problem into separate components and use trigonometry.
1. First, let's calculate the force exerted on the pivot of the crane's arm. We can consider the arm as a lever. The weight of the crane's arm acts as a downward force at a distance of 10.0m from the pivot. Given that the mass of the arm is 1000kg, we can find the force exerted on the pivot using the formula:
Force on the pivot = mass of the arm x acceleration due to gravity
Force on the pivot = (1000kg) x (9.8m/s²)
Force on the pivot = 9800N
2. Now, let's determine the tension in cable A. To do this, we need to consider the forces acting on the block of metal. There are two forces acting on the block: the weight of the block and the tension in cable A.
Let's consider the left side of the arm first. The angle between the cable and the arm is 10°. We can use the sine function to find the vertical component of the tension in cable A:
Vertical component of the tension on the left side = tension in cable A x sin(angle)
Vertical component of the tension on the left side = Tension in cable A x sin(10°)
Considering the right side of the arm, the angle between the cable and the block is 45°. We can use the cosine function to find the vertical component of the tension in cable A:
Vertical component of the tension on the right side = tension in cable A x cos(angle)
Vertical component of the tension on the right side = Tension in cable A x cos(45°)
Since the vertical components of the tension on both sides must balance the weight of the block, we can set up an equation:
Vertical component of the tension on the left side + Vertical component of the tension on the right side = weight of the block
Tension in cable A x sin(10°) + Tension in cable A x cos(45°) = weight of the block
Tension in cable A x [sin(10°) + cos(45°)] = weight of the block
The weight of the block is given as 10,000kg x 9.8m/s² = 98,000N.
Thus, we can solve for the tension in cable A:
Tension in cable A = weight of the block / [sin(10°) + cos(45°)]
Tension in cable A = 98,000N / [sin(10°) + cos(45°)]
Now, to answer your question about when to use sine or cosine:
- Sine (sin) is used when you have an angle and you want to find the vertical component of a force or vector.
- Cosine (cos) is used when you have an angle and you want to find the horizontal component of a force or vector.
In this case, we used sine to find the vertical component of tension on the left side of the arm and cosine to find the vertical component of tension on the right side.