# Newton's Law and Friction (Ramp problem)

In summary, a block is placed on an inclined plane with a 21 N force pushing it up at an angle of 13 degrees. The plane has an angle of 21 degrees and a coefficient of friction of 0.17. The y-component of the force pushing the block up is equal to the weight of the block, while the x-component is equal to the friction force. After solving for mg, the answer was incorrect due to using the wrong angle for the force pushing the block up the ramp. The correct angle should be used to determine the normal force and friction force on the block.

## Homework Statement

A block is placed on a incline plane. A 21 N force is required to push the block up the incline with constant velocity. What is the weight of the block? Answer in units of N

## Homework Equations

The force pushing the block=21 N with an angle of 13 degrees. The ramp has an angle of 21 degrees. Coefficient of friction=0.17. Ramp goes up from left to right.
fs = uN, F = ma, acceleration is 0.

Angle of force = phi, Angle of ramp = theta

## The Attempt at a Solution

My attempt: (after drawing my free-body diagram)

F(y-component): N-mgcos(theta) = 0 --> N=mgcos(theta) --> Fsin(phi)-mgcos(theta) = 0
F(x-component): Fcos(phi)-mgsin(theta)-fs = 0 --> Fcos(phi)-mgsin(theta)-umgcos(theta) = 0

I multiplied the y-component with the negative u (or mu) to cancel out umgcos(theta)

Then: Fcos(phi)-uFsin(phi)-mgsin(theta) = 0, So I solved for mg and my answer was incorrect. What am I doing wrong? Any help appreciated, thanks.

What did you find when the solving was done?
I notice phi is different from the 21 degrees. Is phi wrt horizontal? Does its y component contribute to the friction force ?

Well the problem indicated that the force pushing the block up the ramp had an angle of 13 degrees. So denoting that phi, I used Fsin(phi) and Fcos(phi) respectively.

My answer in the end was about 54.85 N. Which was incorrect. No idea what I am doing wrong.

I take it the free body diagram decomposes mg along the slope and perpendicular to it.
You want to do the same with the 21 N force. What is the angle wrt the plane ?
The component perpendicular to the plane is to be added to mg ##\cos\theta## to determine the normal force.
That times ##\mu## is the friction force, along the slope, downwards. To be added to mg ##\sin\theta##, also downwards. Call the sum Fdown
No acceleration (i.e. constant speed) along the slope means the 21 N component along the slope is equal and opposite to this Fdown

1 person

I can confirm that your approach is correct. However, it seems like you made a small mistake in your calculations. When you multiplied the y-component with the coefficient of friction (u), you forgot to include it in the x-component as well. The correct equation should be: Fcos(phi)-uFsin(phi)-mgsin(theta)-umgcos(theta) = 0. This should give you the correct answer for the weight of the block. Keep up the good work!

## 1. What is Newton's First Law and how does it relate to the ramp problem?

Newton's First Law states that an object at rest will remain at rest and an object in motion will continue in a straight line at a constant speed unless acted upon by an external force. This law is relevant to the ramp problem because it explains why objects on a ramp will stay in place or continue rolling down at a constant speed unless a force, such as friction, is acting on them.

## 2. How does friction affect the motion of an object on a ramp?

Friction is a force that acts in the opposite direction of an object's motion. In the ramp problem, friction can slow down or stop an object from sliding down the ramp. This is because the rough surface of the ramp creates resistance against the object's motion, causing it to slow down or stop.

## 3. How does the angle of the ramp affect the amount of friction?

The steeper the angle of the ramp, the greater the force of friction acting on the object. This is because a steeper ramp creates a larger contact area between the object and the surface, resulting in more friction. Therefore, the angle of the ramp can significantly impact the amount of friction and the motion of the object on the ramp.

## 4. Can friction ever be beneficial in the ramp problem?

Yes, friction can be beneficial in certain scenarios of the ramp problem. For example, in a car going up a steep ramp, friction between the wheels and the ramp helps to provide traction and prevent the car from sliding back down. Additionally, friction can also be used to control the speed of an object on a ramp, such as in a roller coaster where brakes are used to create friction and slow down the cars.

## 5. How can we reduce the effects of friction in the ramp problem?

To reduce the effects of friction in the ramp problem, we can use smoother surfaces or lubricants (such as oil or grease) to make the contact between the object and the ramp smoother. We can also decrease the angle of the ramp, as a smaller contact area will result in less friction. Additionally, using materials with less roughness can also help to reduce friction.

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