# Find when the acceleration is 0

## 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 dont 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 dont know where to start solving this one. Help please!!

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jbriggs444
Homework Helper
As far as I know, acceleration is zero when the forces in the x direction are zero
What forces act in the x direction?

What forces act in the x direction?
kinetic friction and tension

jbriggs444
Homework Helper
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?

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

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
Homework Helper
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?

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

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
Homework Helper
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
Homework Helper
Gold Member
2020 Award
the angle came from using the length and height given in the problem
Doesn't it depend on x?

Doesn't it depend on x?
yes that is what i meant

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
Homework Helper
Gold Member
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 17.3o, you also need a horizontal length. So the question is, what is that horizontal length and how did you get it?

haruspex