Find when the acceleration is 0

[alexi_b]

Homework Statement

A block of mass 2.42kg is accelerated across a rough surface by a rope passing over a pulley, as shown in the figure below. The tension in the rope is 13.9N, and the pulley is 12.3cm above the top of the block. The coefficient of kinetic friction is 0.395. (a: 2.28m/s^2)
Calculate the value of x at which the acceleration becomes zero

(See the attached file for an image of the problem)
2. Homework Equations

The Attempt at a Solution

I don't know where to start with this one. As far as I know, acceleration is zero when the forces in the x direction are zero but i don't know where to start solving this one. Help please!

 

[jbriggs444]

As far as I know, acceleration is zero when the forces in the x direction are zero

What forces act in the x direction?

 

[alexi_b]

What forces act in the x direction?

kinetic friction and tension

 

[jbriggs444]

kinetic friction and tension

Does all of the tension act in the x direction?
How would you go about calculating the frictional force?
In which directions do the two forces act?

 

[alexi_b]

Does all of the tension act in the x direction?
How would you go about calculating the frictional force?
In which direction do the two forces act?

It would be Fnet = Tcos(theta) - kinetic friction (as they act in opposite directions)
then for calculating frictional for I would use F(kinetic)=u*N

right?

 

[jbriggs444]

Right. So your next challenge is calculating the normal force (N).

 

[alexi_b]

Right. So your next challenge is calculating the normal force (N).

right so i got 23.1 N, but I still have no idea how to continue this problem

 

[jbriggs444]

right so i got 23.1 N, but I still have no idea how to continue this problem

How did you arrive at 23.1 N? What forces act in the y direction?

 

[alexi_b]

How did you arrive at 23.1 N? What forces act in the y direction?

Oh right! Sorry i calculated too quickly
Fnet = Fn + Tsin(theta) - Fg
0 = Fn + Tsin(theta) - Fg
Fg - Tsintheta = Fn
23.1 N - Tsin17.3 degrees = Fn
23.1 - 0.297T = Fn

 

[alexi_b]

Oh right! Sorry i calculated too quickly
Fnet = Fn + Tsin(theta) - Fg
0 = Fn + Tsin(theta) - Fg
Fg - Tsintheta = Fn
23.1 N - Tsin17.3 degrees = Fn
23.1 - 0.297T = Fn

the angle came from using the length and height given in the problem

 

[jbriggs444]

the angle came from using the length and height given in the problem

No length was given in the problem. The horizontal displacement between pulley and mass is what you are asked to determine.

 

[haruspex]

the angle came from using the length and height given in the problem

Doesn't it depend on x?

 

[alexi_b]

Doesn't it depend on x?

yes that is what i meant

 

[alexi_b]

No length was given in the problem. The horizontal displacement between pulley and mass is what you are asked to determine.

yeah but i used those measurements to calculate the angle at which i was pulling at, no?

 

[kuruman]

yeah but i used those measurements to calculate the angle at which i was pulling at, no?

The point made by @jbriggs444 is that you are given only one length, the vertical height of the pulley above the top of the block. To obtain an angle of 17.3^o, you also need a horizontal length. So the question is, what is that horizontal length and how did you get it?

 

[haruspex]

i used those measurements

What measurements? You cannot find that angle without knowing x, and x is the value you are to find.
What equation relates x to the angle?