1982 Physics B Question 2 help

[iluvschool699]

A crane is used to hoist a load of mass m1=500 kilograms. The Load is suspended by a cable from a hook of mass m2=50 kilograms, as shown in the diagram above. The load is lifted upward at a constant acceleration of 2 m/s^2.

a) On the diagrams below draw and label the forces acting on the hook and the forces on the load as they accelerate upward.


b)Determine the Tension T1 in the lower cable and the tension T2 in the upper cable as the hook and load are accelerated upward at 2 m/s^2. Use g=10 m/s^2.

 

[Nate810]

T1 = m1 * g + (m1 + m2) * a = 500 * 10 + (500 + 50) * 2 = 10500 NT2 = m2 * g + (m1 + m2) * a = 50 * 10 + (500 + 50) * 2 = 1050 N

 

[ogb p]

I would approach this question by first analyzing the given information and understanding the scenario. The problem describes a crane lifting a load with a constant acceleration of 2 m/s^2. The load has a mass of 500 kilograms and is suspended by a cable attached to a hook with a mass of 50 kilograms. The problem also provides the value of acceleration due to gravity, g, as 10 m/s^2.

a) To answer part a of the question, we need to draw and label the forces acting on the hook and the load as they accelerate upward. From our understanding of Newton's Second Law, we know that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force acting on the load and the hook must be equal to the product of their respective masses and the acceleration of 2 m/s^2. This force is provided by the tension in the cables attached to the hook. Therefore, the forces acting on the hook and the load can be represented by the following free-body diagrams:

Hook: The forces acting on the hook are its weight (mg) and the tension in the upper cable (T2). Since the hook is accelerating upward, the net force acting on it is in the upward direction.

Load: The forces acting on the load are its weight (mg), the tension in the lower cable (T1), and the tension in the upper cable (T2). Since the load is also accelerating upward, the net force acting on it is in the upward direction.

b) To determine the tension in the lower cable (T1) and the upper cable (T2), we can use the equation F=ma, where F is the net force, m is the mass, and a is the acceleration. Since we know the net force acting on the hook and the load, we can calculate the tension in the cables as follows:

T1 = m1a + mg = (500 kg)(2 m/s^2) + (500 kg)(10 m/s^2) = 6000 N
T2 = m2a + mg = (50 kg)(2 m/s^2) + (50 kg)(10 m/s^2) = 600 N

Therefore, the tension in the lower cable is 6000 N and the tension in the upper cable is 600 N.

In conclusion, the crane is able to lift the load