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.
Thank you
To solve this problem, we can break it down into two parts: finding the tension in cable A and determining the force exerted on the pivot of the crane's arm.
1. Finding the tension in cable A:
Let's consider the forces acting on the crane's arm. We have the weight of the crane's arm itself (1000 kg) acting downward and the tension in cable A acting upward.
Using the principle of equilibrium, we know that the sum of all the forces acting on the arm must be zero. This gives us the equation:
Tension in cable A - Weight of the arm = 0
Weight of the arm = mass x gravitational acceleration
Weight of the arm = 1000 kg x 9.8 m/s^2 (approximating Earth's gravitational acceleration)
Now, we can solve for the tension in cable A:
Tension in cable A = Weight of the arm
= 1000 kg x 9.8 m/s^2
= 9800 N
Therefore, the tension in cable A is 9800 Newtons.
2. Determining the force exerted on the pivot of the crane's arm:
Now, let's consider the forces acting on the pivot of the crane's arm. We have the tension in cable A acting upward, the weight of the arm acting downward, and the weight of the metal block acting downward.
Using the principle of equilibrium again, we know that the sum of all the forces acting on the pivot must be zero. This gives us the equation:
Force on pivot - Weight of the arm - Weight of the metal block = 0
Weight of the metal block = mass x gravitational acceleration
Weight of the metal block = 10000 kg x 9.8 m/s^2
Now, we can solve for the force exerted on the pivot of the crane's arm:
Force on pivot = Weight of the arm + Weight of the metal block
= 1000 kg x 9.8 m/s^2 + 10000 kg x 9.8 m/s^2
= 10000 kg x 9.8 m/s^2 + 100000 kg x 9.8 m/s^2
= 100000 N + 980000 N
= 1080000 N
Therefore, the force exerted on the pivot of the crane's arm is 1,080,000 Newtons.
Please note that in this answer, we have assumed that the arm and the cable are massless, and we have neglected any friction or other external forces.