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

AI Thread Summary
To determine the force required to slide a 326-N trunk down a 21.0° incline at constant velocity, the components of gravitational force and friction must be analyzed. The weight's components parallel and perpendicular to the incline are calculated as approximately 115.82 N and 304.34 N, respectively. The normal force is derived from these components, leading to a frictional force of about 142.59 N. The net force equation reveals that the applied force must counteract both the gravitational pull and friction to maintain constant velocity. Ultimately, the calculated force needed to achieve this balance is approximately 161.74 N directed up the incline.
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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
 
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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.)
 
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