# How to Calculate Reversed Force of a Wedge | Friction and Wedges Homework

• OmniNewton
In summary: Have a good night!In summary, the problem involves determining the reversed force of P needed to pull out wedge A, given a diagram and information about the coefficients of static friction between various surfaces. By setting up equations for the forces acting on wedge A and solving for the necessary variables, the correct answer for P was found to be 106 lb. The mistake in the initial attempt was a sign error in one of the equations. The expert helper provided guidance and the mistake was corrected.
OmniNewton

## Homework Statement

Provided the following diagram determine the reversed force of P needed to pull out the wedge, A.

## Homework Equations

Given the coefficient of static friction between A and C and between B and D is 0.2 and between A and B static friction is 0.1. Weights of wedges are neglected.
Force of friction = static coefficient * normal (When impending motion is occurring)

## The Attempt at a Solution

Let NC be the normal at C
ND be the normal at D
NB be the normal between wedge A and wedge B
FC be the friction force between wedge A and wall C
FB be the friction force between wedge A and wedge B
FD be the friction force between wedge B and wall D

For FBD of wedge A:

X-direction:
0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)
Y-direction:
Nc - NBcos15 + 0.1NBcos15=0
simplifying Nc=0.94 NB (Equation 2)

For FBD of wedge B:[/B]

X-direction:
NB= ND/(sin15) (Equation 3)

Y-direction:

NBcos15 - 3000 + 0.2ND = 0 (Equation 4)

substituting equation 3 into 4 and solving ND = 763lb

From equation 3 then NB = 2947.9 lb

From Equation 2 NC = 2771lb

From Equation 1 P = 493.5 lb

My answer is considerably off the correct answer which is 106 lb

Last edited:
OmniNewton said:

## Homework Statement

Provided the following diagram determine the reversed force of P needed to pull out the wedge, A.

## Homework Equations

Given the coefficient of static friction between A and C and between B and D is 0.2 and between A and B static friction is 0.1. Weights of wedges are neglected.
Force of friction = static coefficient * normal (When impending motion is occurring)

## The Attempt at a Solution

Let NC be the normal at C
ND be the normal at D
NB be the normal between wedge A and wedge B
FC be the friction force between wedge A and wall C
FB be the friction force between wedge A and wedge B
FD be the friction force between wedge B and wall D

For FBD of wedge A:

X-direction:
0 = 0.2 NC - NBsin15 - 0.1NBcos15
simplifying 0.2 NC - 0.3554NB - P (Equation 1)
Y-direction:
Nc - NBcos15 + 0.1NBcos15=0
simplifying Nc=0.94 NB (Equation 2)

For FBD of wedge B:[/B]

X-direction:
NB= ND/(sin15) (Equation 3)

Y-direction:

NBcos15 - 3000 + 0.2ND = 0 (Equation 4)

substituting equation 3 into 4 and solving ND = 763lb[/B]

From equation 3 then NB = 2947.9 lb

From Equation 2 NC = 2771lb

From Equation 1 P = 493.5 lb

My answer is considerably off the correct answer which is 106 lb
(Try to not use Bold type face except where necessary.)

Isn't P supposed to be pulling the wedge out? That would mean that friction forces acting on wedge A are exerted toward the right.[/B]

SammyS said:
(Try to not use Bold type face except where necessary.)

Isn't P supposed to be pulling the wedge out? That would mean that friction forces acting on wedge A are exerted toward the right.

This is why the question says the reverse force of P so -P. I accommodated for this in my equations. Sorry about the bold my bad

OmniNewton said:
This is why the question says the reverse force of P so -P. I accommodated for this in my equations. Sorry about the bold my bad
OmniNewton said:
X-direction:
0 = 0.2 NC - NBsin15 - 0.1NBcos15
simplifying 0.2 NC - 0.3554NB - P (Equation 1)

Equation 1 is not an equation. There's no " = " in it.

In the line before that there is no P, and shouldn't 0.1NBcos15° be to the right, so have " + " ?

SammyS said:
Equation 1 is not an equation. There's no " = " in it.

In the line before that there is no P, and shouldn't 0.1NBcos15° be to the right, so have " + " ?
Sorry I forgot to add that variable when typing the corrections have been made,
0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)

OmniNewton said:
Sorry I forgot to add that variable when typing the corrections have been made,
0 = 0.2 NC - NBsin15 - 0.1NBcos15 - P
simplifying 0= 0.2 NC - 0.3554NB - P (Equation 1)
That doesn't address the sign on 0.1NBcos15° .

SammyS said:
That doesn't address the sign on 0.1NBcos15° .
Sorry you are absolutely correct let me work through this with this change, thank you!

Excellent, Thank you so much for your time it worked out! Sorry for wasting your time on such a trivial mistake. I was looking at this for hours and didn't notice it.

OmniNewton said:
Excellent, Thank you so much for your time it worked out! Sorry for wasting your time on such a trivial mistake. I was looking at this for hours and didn't notice it.
No problem.

Lately I've been helping some people who understand almost nothing and at times seem to act clueless on purpose. It's great to be able to make a few suggestions and have them recognized for what they are,

SammyS said:
No problem.

Lately I've been helping some people who understand almost nothing and at times seem to act clueless on purpose. It's great to be able to make a few suggestions and have them recognized for what they are,

Yeah I really appreciate that I think it is important to work from ones mistakes and learn from them. I appreciate you not just giving me the answer but rather guiding me.

SammyS

## 1. What is friction and how does it affect wedges?

Friction is the force that resists motion between two surfaces in contact. When a wedge is pushed between two surfaces, friction can either help or hinder its movement depending on the direction of the force applied.

## 2. How does the angle of a wedge affect friction?

The angle of a wedge, also known as its slope, can greatly affect the amount of friction between the wedge and the surfaces it is being pushed against. A steeper angle will result in more friction, while a shallower angle will decrease the amount of friction.

## 3. Can wedges be used to reduce friction?

Yes, wedges can be used to reduce friction in certain situations. For example, a lubricated wedge can reduce friction between two surfaces, making it easier to slide or move objects.

## 4. How does the weight of a wedge affect its friction?

The weight of a wedge can affect its friction in a couple of ways. First, a heavier wedge will have a greater force pushing it into the surfaces, resulting in more friction. Additionally, the weight distribution of a wedge can also impact its friction, as a wedge with more weight towards the front will have more friction than one with more weight towards the back.

## 5. What are some real-world applications of friction and wedges?

Friction and wedges have numerous real-world applications, such as in construction, where wedges are used to split materials like wood and stone. They are also used in tools like knives and scissors to cut through materials. Additionally, friction and wedges are important in transportation, as tires use friction to grip the road and wedges are used in braking systems to slow down vehicles.

• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
4K
• Introductory Physics Homework Help
Replies
13
Views
4K
• Mechanics
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
20
Views
3K