Solving a pulley system (Dynamics)

[mmanor]

Solving a problem about a pulley system (dynamics)

A 100 kg crate is supported by a rope which passes over a 25 kg pulley. A 2nd rope connects the pulley to the ceiling. The system is stationary.

ƩFx=0 ƩFy=0

I am having problems setting up the free body diagram to start the problem!

 

[haruspex]

Where does the other end of the first rope go? Are both ends attached to the crate?

 

[TamuNinja]

well using the information that you gave which is not very clear.

Your freebody diagram would be:

mg acting downward on the 1000kg block and Tension (T) acting upward on the block

Thats just for the block.


There are no foces acting in the X axis

ƩF_y=o=T-mg (this is the sum of forces acting on the block)


Now the info that you gave in your original post was not very clear.. Please correct me if i missed anything and i will go back and make sure i correct everything.

 

[aralbrec]

You haven't stated what you are trying to find. It sounds like you want to find the tension in the rope connected to the ceiling. In that case you need a free body where that tension appears.

 

[Nickie]

I would approach this problem by first identifying the key variables and principles involved. In this case, the variables are the weight of the crate (100 kg), the weight of the pulley (25 kg), and the tension in the two ropes. The principles involved are Newton's laws of motion and the concept of equilibrium.

Next, I would draw a free body diagram, which is a visual representation of all the forces acting on the system. In this case, there are three forces to consider: the weight of the crate acting downwards, the tension in the rope supporting the crate acting upwards, and the tension in the rope connecting the pulley to the ceiling acting upwards. By drawing these forces as vectors, I can see that the system is in equilibrium, as the forces are balanced in both the horizontal and vertical directions.

From here, I would apply Newton's second law (ƩF=ma) to each of the ropes separately. Since the crate is stationary, the sum of forces acting on it must be zero. This means that the tension in the rope supporting the crate must be equal to the weight of the crate (100 kg x 9.8 m/s^2 = 980 N). Similarly, the tension in the rope connecting the pulley to the ceiling must be equal to the combined weight of the crate and the pulley (125 kg x 9.8 m/s^2 = 1225 N).

In summary, solving a problem about a pulley system (dynamics) involves identifying the variables and principles involved, drawing a free body diagram, and applying relevant equations to determine the unknown quantities. By following this systematic approach, we can effectively solve complex problems and gain a deeper understanding of the underlying dynamics at play.