Simple Free Body Force Diagram + Friction Question

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Homework Help Overview

The discussion revolves around a physics problem involving a free body force diagram, focusing on forces acting on a body accelerating up an incline. The body has a mass of 6000 kg and is experiencing an acceleration of 6.34 m/s², with a component of gravitational force acting down the incline.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of net forces acting on the body, including the effects of friction and gravitational components. There is a focus on determining whether sufficient information is available to calculate the force of friction or the coefficient of friction.

Discussion Status

The discussion is ongoing, with participants questioning the terminology used and the approach to identifying all forces acting on the body. Some guidance has been provided regarding the application of Newton's laws and the need to differentiate between cases with and without friction. Participants express uncertainty about solving for the friction force due to having multiple unknowns.

Contextual Notes

Participants note the complexity introduced by considering both frictional and non-frictional scenarios, leading to confusion about the net forces and the relationships between them. There is acknowledgment of the need for an applied force to overcome gravitational effects and accelerate the block.

Procrastinate
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I just want to check if I am doing the right thing with this force diagram.

Forgetting the force of friction for the moment, if I know that the body is accelerating at 6.34ms^-2 with a mass of 6000kg; and the opposing force of mgsin3 is 3077.4N, then would the I have to add that to the acceleration force? Therefore Force acceleration would be 38040 + 3077.4 = 41117.4N...


My actual question is about friction. Do I have sufficient values to calculate the Force of friction or the friction co-efficient?

F of Normal = mu x N
 

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Procrastinate said:
I just want to check if I am doing the right thing with this force diagram.

Forgetting the force of friction for the moment, if I know that the body is accelerating at 6.34ms^-2 with a mass of 6000kg; and the opposing force of mgsin3 is 3077.4N, then would the I have to add that to the acceleration force? Therefore Force acceleration would be 38040 + 3077.4 = 41117.4N...


My actual question is about friction. Do I have sufficient values to calculate the Force of friction or the friction co-efficient?

F of Normal = mu x N
Your terminology for 'acceleration force is a bit unorthodox. It is most always best to identify all real forces acting on a body, determine the NET of those forces in the x and y direction, and apply Newton's laws in that direction (F_net =ma, which, for the x direction, is F_x_net = 38040 ). So what are all the forces acting in the x direction (including friction)? Do you have enough info to solve for the friction force?
 
PhanthomJay said:
Your terminology for 'acceleration force is a bit unorthodox. It is most always best to identify all real forces acting on a body, determine the NET of those forces in the x and y direction, and apply Newton's laws in that direction (F_net =ma, which, for the x direction, is F_x_net = 38040 ). So what are all the forces acting in the x direction (including friction)? Do you have enough info to solve for the friction force?


I thought F_x_net would be 41117.4N.

Um, I don't think so.
 
Procrastinate said:
I thought F_x_net would be 41117.4N.

Um, I don't think so.
No, and you are looking at 2 cases, with and without friction, so let's not confuse the two. When you look at the problem as if it were written without friction, the net force is 38040, which is made up of the applied pulling force acting up the incline, and the weight component acting down the incline. Thus
F_net =ma = 38040
F_applied - mgsin3 = 38040
F_applied - 3077 = 38040
F_applied = 41117 N
which i think is what you were trying to say before; but you were missing the concept of an applied force (by a person, or machine, etc.) required up the plane inorder to overcome gravity and accelearte the block.

Now for the real problem with friction, use the same approach and respond to your own question> do you have enough info to solve for the friction force?
 
PhanthomJay said:
No, and you are looking at 2 cases, with and without friction, so let's not confuse the two. When you look at the problem as if it were written without friction, the net force is 38040, which is made up of the applied pulling force acting up the incline, and the weight component acting down the incline. Thus
F_net =ma = 38040
F_applied - mgsin3 = 38040
F_applied - 3077 = 38040
F_applied = 41117 N
which i think is what you were trying to say before; but you were missing the concept of an applied force (by a person, or machine, etc.) required up the plane inorder to overcome gravity and accelearte the block.

Now for the real problem with friction, use the same approach and respond to your own question> do you have enough info to solve for the friction force?

F_net = 38040 = ma
F_applied - mgsin3 - F_friction = 38040
F_applied - F_friction = 41117N
F_applied - F_normal*mu = 41117N

I have two unknown variables now.
 
Procrastinate said:
F_net = 38040 = ma
F_applied - mgsin3 - F_friction = 38040
F_applied - F_friction = 41117N
F_applied - F_normal*mu = 41117N

I have two unknown variables now.
yes, good. So can you solve for the friction force or not??
 
PhanthomJay said:
yes, good. So can you solve for the friction force or not??

I would say no.
 
Procrastinate said:
I would say no.
And so would I:cry:
 

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