# Pushing at a downward angle with friction

• casedogg
In summary: The first equation I found was mg+Fsin(40) which equals 313.6N. The second equation I found was Fcos(40) which equals 129.5N. So by solving for F, I got 129.5N.
casedogg

## Homework Statement

A factory worker pushes a 32.0 kg crate a distance of 5.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.25. The factory worker pushes downward at an angle of 40° below the horizontal.
What magnitude of force must the worker apply to move the crate at a constant velocity?

Fn= mg+Fy
fk=.25Fn
Fy=Fsin(40)

## The Attempt at a Solution

I spent a long time trying to figure this out, I know mg=313.6N and i need to add the force of the y component to that to find the normal force, but I don't know how to find Fy, or Fx. If there was no vertical push it would just be 78.4N in the x direction, which equals the force of friction. But with the vertical push it seems like the harder you push, the more friction increases, and thus you need to push harder. I saw other people have had the same question on this forum before, but with different numbers, but they don't help me because people only told them that the vertical push makes the normal force bigger, but I don't know how to calculate the vertical push.

Last edited:
casedogg said:

## Homework Statement

A factory worker pushes a 32.0 kg crate a distance of 5.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.25. The factory worker pushes downward at an angle of 40° below the horizontal.
What magnitude of force must the worker apply to move the crate at a constant velocity?

Fn= mg+Fy
fk=.42Fn
Fy=Fsin(40)

## The Attempt at a Solution

I spent a long time trying to figure this out, I know mg=313.6N and i need to add the force of the y component to that to find the normal force, but I don't know how to find Fy, or Fx. If there was no vertical push it would just be 131.712N in the x direction, which equals the force of friction. But with the vertical push it seems like the harder you push, the more friction increases, and thus you need to push harder. I saw other people have had the same question on this forum before, but with different numbers, but they don't help me because people only told them that the vertical push makes the normal force bigger, but I don't know how to calculate the vertical push.

Would you believe that you have already written the equation that gives it to you?
Before I tell you where it is, do you know how to break vectors into components? You use trig. --you know sines and cosines hypotenuses and opposite and adjacent sides and all that good stuff

Up above you wrote $$F_y = F sin(40)$$ It seems you knew how to do it all along...

But I don't know what Fy is or F, what I'm trying to find is F, if I could find Fy, Fx, fk or Fn then I would be able to find F

casedogg said:
But I don't know what Fy is or F, what I'm trying to find is F, if I could find Fy, Fx, fk or Fn then I would be able to find F

Well, first since the velocity is constant, the acceleration is zero and so the sum of the forces in the x direction is zero. That implies that the frictional force is equal in magnitude to the applied horizontal force or,

$$f = F cos(40)$$

But,as you know, the frictional force is also given by

$$f = \mu_k(mg + Fsin(40) )$$

You know $$\mu_k$$ and m and g and you can set those two equations equal and get F.

UniqueName
F=129.5N
Thank You very much!

I had been trying to find two equations like that which equaled each other so I would only have one variable.

## 1. What is the concept of pushing at a downward angle with friction?

The concept of pushing at a downward angle with friction is the act of applying force on an object in a downward direction while also exerting a force in the opposite direction to counteract the frictional force between the object and the surface it is resting on.

## 2. How does pushing at a downward angle affect the frictional force?

Pushing at a downward angle increases the frictional force between the object and the surface. This is because the downward force creates a normal force between the two surfaces, which in turn increases the frictional force.

## 3. What factors affect the effectiveness of pushing at a downward angle with friction?

The effectiveness of pushing at a downward angle with friction is affected by the magnitude of the downward force, the coefficient of friction between the two surfaces, and the angle at which the force is applied.

## 4. Can pushing at a downward angle with friction be useful in real-life situations?

Yes, pushing at a downward angle with friction is commonly used in various real-life situations, such as when pushing a heavy object on a flat surface or when using a handbrake in a car to prevent it from rolling down a slope.

## 5. How can pushing at a downward angle with friction be calculated and measured?

To calculate the force required to push an object at a downward angle with friction, you can use the formula F = μN, where F is the force, μ is the coefficient of friction, and N is the normal force. This force can be measured using a force meter or by calculating the weight of the object and multiplying it by the coefficient of friction.

• Introductory Physics Homework Help
Replies
42
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
5K
• Introductory Physics Homework Help
Replies
38
Views
2K
• Introductory Physics Homework Help
Replies
41
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
2K