# Find when the acceleration is 0

• alexi_b
In summary: 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?
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!

#### Attachments

• Screenshot (11).png
5.6 KB · Views: 374
alexi_b said:
As far as I know, acceleration is zero when the forces in the x direction are zero
What forces act in the x direction?

jbriggs444 said:
What forces act in the x direction?
kinetic friction and tension

alexi_b said:
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?

jbriggs444 said:
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?

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

jbriggs444 said:
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

alexi_b said:
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?

jbriggs444 said:
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 said:
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

alexi_b said:
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.

alexi_b said:
the angle came from using the length and height given in the problem
Doesn't it depend on x?

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

jbriggs444 said:
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?

alexi_b said:
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?

alexi_b said:
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?

## 1. What does it mean for acceleration to be 0?

When acceleration is 0, it means that there is no change in the velocity of an object. This could either mean that the object is moving at a constant speed, or it is completely still.

## 2. How do you find when the acceleration is 0?

To find when the acceleration is 0, you need to analyze the motion of the object over a specific time interval. This can be done by measuring the object's velocity at different points in time and calculating the change in velocity over that time period. If the change in velocity is 0, then the acceleration is also 0.

## 3. What factors can cause acceleration to be 0?

Acceleration can be 0 due to two main factors: constant velocity or no motion. In the case of constant velocity, the object is moving at a steady rate without any change in speed or direction. In the case of no motion, the object is completely still and not moving at all.

## 4. Why is it important to find when acceleration is 0?

Finding when acceleration is 0 can provide valuable information about the motion of an object. It can help determine if the object is moving at a constant speed, if there is any change in the direction of motion, and if there are any external forces acting on the object.

## 5. How does knowing when acceleration is 0 relate to Newton's Laws of Motion?

Newton's Laws of Motion state that an object will remain at rest or in a state of constant velocity unless acted upon by an external force. Therefore, if acceleration is 0, it means that there is no net force acting on the object, which is in line with Newton's First Law. Additionally, Newton's Second Law states that force is equal to mass multiplied by acceleration, so if acceleration is 0, then there must be no net force acting on the object.

• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
776
• Introductory Physics Homework Help
Replies
2
Views
669
• Introductory Physics Homework Help
Replies
38
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
244
• Introductory Physics Homework Help
Replies
22
Views
3K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
957
• Introductory Physics Homework Help
Replies
4
Views
979
• Introductory Physics Homework Help
Replies
3
Views
1K