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

In summary: The weight's y component (magnitude) is:(weight) + (friction) = 0The friction's y component is:(friction) = -.322
  • #1
halo9909
37
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
 
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  • #2
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?
 
  • #3
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,
 
  • #4
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?
 
  • #5
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
 
  • #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.)
 

1. What is force friction at an angle?

Force friction at an angle refers to the resistance or opposition to motion that occurs when two surfaces are in contact with each other at an angle, rather than parallel to each other. This type of friction is also known as "sliding friction" or "tangential friction."

2. How is force friction at an angle different from regular friction?

The main difference between force friction at an angle and regular friction is the direction of the force. Regular friction is parallel to the surfaces in contact, while force friction at an angle is perpendicular to the surfaces. This means that force friction at an angle is always smaller than regular friction, as it is only a component of the total friction force.

3. What factors affect force friction at an angle?

The factors that affect force friction at an angle include the type of surfaces in contact, the roughness of the surfaces, the amount of force applied, and the angle at which the surfaces are in contact. The coefficient of friction, which is a measure of how easily two surfaces slide against each other, also plays a role in force friction at an angle.

4. How is force friction at an angle calculated?

The force of friction at an angle can be calculated using the formula: Ff = µ * Fn * cos(θ), where Ff is the force of friction, µ is the coefficient of friction, Fn is the normal force (perpendicular force) between the surfaces, and θ is the angle at which the surfaces are in contact. This formula assumes that the surfaces are in motion and that the force of friction is constant.

5. How can force friction at an angle be reduced?

Force friction at an angle can be reduced by using a lubricant, such as oil or grease, between the surfaces. This reduces the coefficient of friction and makes it easier for the surfaces to slide against each other. Additionally, using smoother or more polished surfaces can also reduce force friction at an angle. Increasing the angle at which the surfaces are in contact can also reduce force friction, as the force of friction is proportional to the cosine of the angle.

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