I'm sorry, I am not sure what you are asking for. Could you clarify?

Click For Summary

Homework Help Overview

The problem involves calculating the force required to move a 326-N trunk up a 21.0° inclined plane at a constant velocity, considering the coefficient of friction between the trunk and the plane is 0.322.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the components of forces acting on the trunk, including weight and friction. There are attempts to break down the forces into parallel and perpendicular components relative to the incline.

Discussion Status

Some participants are exploring the equations of motion in both the vertical and horizontal directions, while others are questioning the calculations of force components and the normal force. There is an ongoing dialogue about the correct application of force equations and the implications of zero acceleration.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on understanding the relationships between different forces rather than arriving at a final answer.

halo9909
Messages
37
Reaction score
0

Homework Statement



What force would you have to exert on a 326-N trunk up a 21.0° inclined plane so that it would slide down the plane with a constant velocity? What would be the direction of the force? (The coefficient of friction between the plane and the trunk is 0.322.)


Homework Equations



not sure what to use by
Ff=coefficient friction *Fn

The Attempt at a Solution


Ihave tried drawing a picture of this
but what i done is
figure that the Fw=326N
from there I am not sure where to use it or what equations to
 
Physics news on Phys.org
Hi halo9909,

halo9909 said:

Homework Statement



What force would you have to exert on a 326-N trunk up a 21.0° inclined plane so that it would slide down the plane with a constant velocity? What would be the direction of the force? (The coefficient of friction between the plane and the trunk is 0.322.)


Homework Equations



not sure what to use by
Ff=coefficient friction *Fn

The Attempt at a Solution


Ihave tried drawing a picture of this
but what i done is
figure that the Fw=326N
from there I am not sure where to use it or what equations to

From your picture, what force components are parallel to the incline? what force components are perpendicular to the incline? Once you have those, you can write two force equations. What do you get?
 
well since 326N is the Fw
so i would use
326sin21=115.82N=Fay
326cos21=304.34=Fax
so Fnormal=326+115.82=442.8
Ff=.322*442.8=142.59 since Ff=mu*Fnormal

then
Fnet= Fa-Ff
Fnet=304.34-142.6
=161.74
From here I am unsure what to do,
 
halo9909 said:
well since 326N is the Fw
so i would use
326sin21=115.82N=Fay

Didn't you get 116.828N for this?

326cos21=304.34=Fax

These are fine; these are the components of the weight force parallel (x) and perpendicular (y) to the incline.

Now you can use this in the force equations. I would start with the perpendicular (y) direction. What forces have components in the y direction? And what is the acceleration in the y direction? Once you have those you use them in:

(sum of all y-components of all forces ) = (mass) (acceleration in y direction)

which is what Fnet,y=may means.

So you should be able to find the normal force from that equation. What do you get for that? (Remember that the components can be positive or negative.)

After you have the normal force, repeat the process for the parallel (x) direction. Using the normal force and the equation for kinetic friction in it, you should be able to find F. What do you get?
 
alphysicist said:
Didn't you get 116.828N for this?



These are fine; these are the components of the weight force parallel (x) and perpendicular (y) to the incline.

Now you can use this in the force equations. I would start with the perpendicular (y) direction. What forces have components in the y direction? And what is the acceleration in the y direction? Once you have those you use them in:

(sum of all y-components of all forces ) = (mass) (acceleration in y direction)

which is what Fnet,y=may means.

So you should be able to find the normal force from that equation. What do you get for that? (Remember that the components can be positive or negative.)

After you have the normal force, repeat the process for the parallel (x) direction. Using the normal force and the equation for kinetic friction in it, you should be able to find F. What do you get?

so the Fnormal for the vertical force would be 442.82=326+116.82
and the horizontal would still be 161.74=394.34-142.6
 
halo9909 said:
so the Fnormal for the vertical force would be 442.82=326+116.82

No; the equation you need to start with is essentially:

(y component of normal force) + (y component of weight) + (y component of friction)+(y component of applied force) = (mass) (acceleration in y direction)

Now they tell you the acceleration is zero, and the normal force is always perpendicular to the plane, so this becomes:

(normal force) + (y component of weight) + (y component of friction)+(y component of F)=0

You have already found the weight's y component (magnitude); what is the y component of the friction and the force F? (It will turn out that finding the normal force is simpler than your expression.)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
10
Views
20K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
3K
Replies
9
Views
5K
Replies
5
Views
2K
Replies
3
Views
4K
Replies
15
Views
2K