What is the Direction and Magnitude of the Force on a Hanging Rope?

AI Thread Summary
The discussion revolves around determining the direction and magnitude of forces acting on a hanging rope and a block influenced by multiple forces. Participants clarify the calculations for acceleration, velocity, and the resultant force, emphasizing the need for initial conditions to solve the problem accurately. The direction of the acceleration is identified as 280 degrees, indicating it is in the fourth quadrant. There is a consensus that the direction of the mass's movement aligns with the resultant force, especially since it starts from rest. Overall, the conversation highlights the importance of understanding vector components in analyzing motion and forces.
Niles
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Homework Statement



1) A block is affected by three forces, and I have found the direction and magnitude of the resulting force, and thus the direction and magnitude of the acceleration. I know that the magnitude of the velocity vector for t = 5 is 4.7 m/s and that the distance traveled for t = 5 s is 12 m.

The question is, in which direction the mass is moving at time t = 5 seconds?

It seems like I need some more info? Or am I missing the big picture here?

2) A rope is hanging from the ceiling, and I have to find the direction of the force that is exerted by the ceiling on the rope. I believe, it is upwards - and the magnitude of m*g of the rope?

Thanks in advance.
 
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Can you post the exact question for the first one? What acceleration did you get?

is t=0 the point when acceleration starts taking effect?

Your answer for the second question looks good.
 
I posted the _exact_ question in 1). I get the acceleration 0.92 m/s^2
 
what's the direction of acceleration?
 
I'm still a little confused about the question. But taking position at t=0 to be x=0, and taking acceleration as 0.92m/s^2, and using x=12 at t=5, solve for v at t=5s.

So use this equation:

x = v0*t + (1/2)at^2

and the velocity equation:

v = v0 + at

So you'll get v magnitude and direction... I'm not sure why they tell you the magnitude... seems like you can just calculate it.
 
Direction of the acceleration is 280 degrees (4. quadrant).

There are some previous questions which gave me those answers, and the last question is what I typed in #1.

Wouldn't direction require for me to have the components of v?
 
Niles said:
Direction of the acceleration is 280 degrees (4. quadrant).

There are some previous questions which gave me those answers, and the last question is what I typed in #1.

Wouldn't direction require for me to have the components of v?

Maybe it's better if you type in all the parts to the question. :wink:
 
My bad.

1) Calculate the magnitude of the total resultant force acting on the mass.

2) What angle does the resultant make with the positive x axis?

3) What is the magnitude of the mass's acceleration vector?

4) What is the direction of the acceleraton-vector? In other words, what angle does this vector make with respect to the positive x axis?

5) How far (in meters) will the mass move in 5.0 s?

6) What is the magnitude of the velocity vector of the block at t = 5?

7) In what direction is the mass moving at time t = 5? That is, what angle does the velocity vector make with respect to the positive x axis?
 
Wouldn't the direction of the velocity just be the direction of the acceleration? Or was there an initial velocity?

Any other information? It would be good if you could post a picture... I'm still unclear about the problem...
 
  • #10
It is initially at rest on the surface, and the forces on the picture are:

F1 = 4N

F2 = 6N

F3 = 8N

I think it has the same direction as the acceleration.
 

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  • #11
Niles said:
It is initially at rest on the surface, and the forces on the picture are:

F1 = 4N

F2 = 6N

F3 = 8N

I think it has the same direction as the acceleration.

I think that it is a problem of linear dynamics. Hence as there is no initial velocity the direction of the mass would be in the direction of the resultant force. More over velocity is in the direction of the resultant force. So as learning said acc is in the direction of velocity.
 
  • #12
Thanks guys.
 
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