Newton's Laws Questions: Solving for Force on a 3kg Block on an Incline

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Homework Help Overview

The discussion revolves around a physics problem involving Newton's laws, specifically focusing on a 3 kg block moving on a 40-degree incline. The original poster seeks to determine the force required for the block to move down the incline at a constant velocity, given that it is currently moving up with a 26-N force.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the forces acting on the block, including the applied force, gravitational force components, and friction. There are questions about the role of friction and the correct interpretation of forces in the context of constant velocity.

Discussion Status

Participants are actively engaging with the problem, offering hints and corrections regarding the calculations and assumptions made by the original poster. There is an ongoing exploration of the necessary components of forces and the implications of moving at constant velocity.

Contextual Notes

Some participants note the importance of understanding the frictional forces involved and the assumption of gravity's acceleration. There is also mention of the need for a clear diagram to visualize the forces acting on the block.

graten2go
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[SOLVED] Newtons Laws Questions

Question: A 3 kg block moves up a 40 degrees incline with constant speed under the action of a 26-N force up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?

Here is what I have done so far:
26 N sin 40 degrees and I got 16.8. However, I think I am doing it wrong. Can someone help me out? Thanks!
 
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i think you need to know the force due to friction on the inclined plane, or the coefficients of sliding friction.

but if the inclined plane is assumed frictionless. there is a quick and easy (but it requires insight) answer to your question.
 
actually, i need to correct myself. there is enough information to compute the forces due to friction, so without any assumption of frictionless, the problem can be solved and it is non-trivial (although it's reasonably easy). i think you do have to assume the acceleration of gravity is 9.8 m/s2 or similar.

set up a drawing of this and remember what must be true of the block if it is moving at at constant velocity.
 
graten2go said:
Question: A 3 kg block moves up a 40 degrees incline with constant speed under the action of a 26-N force up and parallel to the incline.
Use this information to figure out the force of friction. Since the block moves up at a constant speed, the forces parallel to the incline must add to what?

Here is what I have done so far:
26 N sin 40 degrees and I got 16.8. However, I think I am doing it wrong.
Since the 26 N force is parallel to the incline, there's no need to find its component in some other direction.

Hint: What's the component of the block's weight parallel to the incline?
 
Can you please offer more of a hint?
I used trig to find the y component of the block, but what do I do with it? Thanks!
 
The 26N force acts parallel to the plane... what is the component of gravity parallel to the plane?

Find the frictional force for the first part... you know the net force parallel to the plane = 0

There are 3 forces involved... the 26N, the component of gravity parallel to the plane, and friction.

Once you get friction, you can do the second part... the frictional force is exactly the same, but it is the opposite direction. Again net force parallel to the plane = 0.
 
graten2go said:
Question: A 3 kg block moves up a 40 degrees incline with constant speed under the action of a 26-N force up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?

Here is what I have done so far:
26 N sin 40 degrees and I got 16.8. However, I think I am doing it wrong. Can someone help me out? Thanks!

you shouldn't multiply 26N by sin40. The 26N is parallel to the incline.
 

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