A large 2,000-N cement block is pulled up a frictionless 43 degrees incline using a dual-strand pulley that is attached to the block. One end of the rope is tied to the wall at the top of the plane while a man is pulling the other end of the rope. What is the tension in the pull of the rope if the rope is pulled parallel to the plane?
(A) 732 N
(B) 682 N
(C) 1,463 N
(D) 1,364 N
(E) 462 N
component of weight down slope =2000 sin 43 = 1364
1364/2 = 682 Newtons
Well, well, well, looks like we've got a physics problem here. Let's tackle it with a dash of humor, shall we?
So, we have a cement block and a man pulling a rope. It's like a tug of war, but with physics involved. Now, imagine this: the cement block is like a sleepy elephant trying to climb up a hill, and the man is like a tiny ant, giving it a helping hand.
Now, for the actual calculation. We need to find the tension in the rope, which depends on the forces acting on the block. Since the block is being pulled parallel to the incline, we need to consider the component of the gravitational force along the incline.
Using a bit of trigonometry, we can find that the component of the gravitational force along the incline is given by Fg * sin(43°), where Fg is the weight of the block.
The weight of the block is given by m * g, where m is the mass of the block (which we don't know) and g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
Now, let's plug in the numbers and solve this beast of a problem:
Weight of the block = 2000 N
sin(43°) = 0.6820 (approximately)
Tension in the rope = 2000 N * 0.6820 = 1364 N.
So, my friend, the answer is (D) 1,364 N. That's the tension in the pull of the rope. The little ant is giving it quite the tug, huh?
To find the tension in the pull of the rope, we can break down the forces acting on the block.
1. Resolve the force of gravity: The force of gravity acting on the block can be resolved into two components. One component acts perpendicular to the plane, and the other component acts parallel to the plane. The perpendicular component is given by the equation F_perpendicular = mg * cos(θ), where m is the mass of the block and θ is the angle of the incline.
In this case, the mass of the block can be calculated using the equation m = F_gravity / g, where F_gravity is the force of gravity acting on the block and g is the acceleration due to gravity. Therefore, m = 2000 N / 9.8 m/s^2 = 204.08 kg.
So, F_perpendicular = 204.08 kg * 9.8 m/s^2 * cos(43°) ≈ 1,363.64 N.
2. Determine the net force and acceleration: To move the block up the incline with constant velocity, the net force acting on the block must be zero. This means that the component of the tension force acting parallel to the incline must balance the component of the force of gravity acting parallel to the incline.
The parallel component of the force of gravity is given by the equation F_parallel = mg * sin(θ), where θ is the angle of the incline.
F_parallel = 204.08 kg * 9.8 m/s^2 * sin(43°) ≈ 1489.54 N.
Since the net force is zero and the force of gravity is acting in the opposite direction to the tension force, the tension force must be equal in magnitude but in the opposite direction.
Therefore, the tension in the pull of the rope is approximately 1489.54 N.
So, the correct answer is (C) 1,463 N.
To find the tension in the pull of the rope, we need to analyze the forces acting on the cement block.
First, we need to resolve the gravitational force acting on the block into its components parallel and perpendicular to the incline. The perpendicular component does not affect the tension in the rope since it is directed into the incline and is balanced by the normal force.
The parallel component of the gravitational force, F_parallel = m * g * sin(θ), where m is the mass of the block and θ is the angle of the incline (43 degrees in this case).
Next, we need to consider the forces acting on the block in the parallel direction. We have the tension force in the rope and the parallel component of the gravitational force. These forces must be equal and opposite to maintain equilibrium.
So, Tension = F_parallel = m * g * sin(θ)
Since we are given the weight of the block as 2,000 N, we can convert it to mass using the formula m = W / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
m = 2000 N / 9.8 m/s^2 = 204.08 kg (rounded to two decimal places)
Substituting the values into the formula, we get:
Tension = 204.08 kg * 9.8 m/s^2 * sin(43 degrees) = 1,463.34 N (rounded to the nearest whole number)
Therefore, the tension in the pull of the rope if the rope is pulled parallel to the plane is approximately 1,463 N.
The correct answer is (C) 1,463 N.