# I need the normal force, the force of friction, and her acceleration

• sheebajenny
In summary, Dorothy has two problems. The first problem is Greg is pushing a 32 kg crate across the floor. He is making it difficult by applying an 800 N force down at an angle of 37 degrees below the horizontal. The coefficient of kinetic friction between the crate and the floor is .50. Dorothy needs to find the normal force, and the acceleration. In the second problem, after completing her jump, a ski jumper is going back up a hill on the other side. The hill is angled at 37 degrees to the horizontal. The coefficient of kinetic friction between her bargain basement skis and the snow is .20. Her mass, including all equipment, is 80 kg. Dorothy needs the normal force, the force of friction, and her acceleration
sheebajenny
I've got 2 problems:
Number 1: Greg is pushing a 32 kg crate across the floor. He is making like difficult by applying his 800 N force down at an angle of 37 degrees below the horizontal. The coefficient of kinetic friction between the crate and the floor is .50.
I need to find the normal force, and the acceleration.
Number 2: After completing her jump, a ski jumper is going back up a hill on the other side, which is designed to slow her down. The hill is angled at 37 degrees to the horizontal. The coefficient of kinetic friction between her bargain basement skis and the snow is .20. Her mass, including all equipment, is 80kg
I need the normal force, the force of friction, and her acceleration

Last edited by a moderator:
Welcome to the PF, Jenny. One of the rules here is that you have to show us some of your work in order for us to help you. What can you tell us about frictional forces? What direction do they act in? What is the equation relating a force F to an object's mass and its resulting acceleration?

here is what I know force of friction=coefficient of friction*Normal force. To find normal force you use SOH CAH TOA depending. They act opposing the force of in the first problem 800N. g*sin(theda)-acceleration/g*cos(theda) acceleration= 2*distance/time squared. I'm really having a difficult time understanding the whole tension concept with friction thrown in. By themselves I can somewhat understand, but when they are combined I got lost.

I may have figured it our #1 normal force sin37=320/N N=531.7
acceleration .5=10sin37-a/10cos37 a=2m/s*s with major rounding
#2 normal force= 800N? force of friction=160N? accleration .2=10sin37-a/10cos37 a=-4.4m/s*s

sheebajenny said:
here is what I know force of friction=coefficient of friction*Normal force. To find normal force you use SOH CAH TOA depending. They act opposing the force of in the first problem 800N. g*sin(theda)-acceleration/g*cos(theda) acceleration= 2*distance/time squared. I'm really having a difficult time understanding the whole tension concept with friction thrown in. By themselves I can somewhat understand, but when they are combined I got lost.

Hi Sheebajenny,

I find drawing a freebody diagram makes life much simpler. Have you done that?

Once you've done that, you can find the net forces in the horizontal and vertical direction.

One hint: the equation you gave, a = 2d/t^2, probably won't be that useful in this problem. Consider using Newton's second law:

Net Force = ma

Where Net force is the sum of all the forces in a certain direction.

Try this, and let us know if you run up against more problems.

Dorothy

I did draw free body diagrams. For the second problem my only force I have is weight it seems like I must be missing something.

correction then I first found acceleration and then I used that to find normal force and then force of friction for the second problem one so I got normal force=325N and force of friction as 70.4N

sheebajenny said:
I may have figured it our #1 normal force sin37=320/N N=531.7
acceleration .5=10sin37-a/10cos37 a=2m/s*s with major rounding
#2 normal force= 800N? force of friction=160N? accleration .2=10sin37-a/10cos37 a=-4.4m/s*s

Are you using g=10 m/s^2?

This is a good try, but I don't think it's quite right yet.

The normal force would be a combination of the downward part of the push and the pull of gravity on the crate.

If you draw a free body diagram, you'll see the normal force pushes up, from the floor, and the other forces (or the components) are straight down. Since there is no motion, they add up to zero:

N - ForceFromPush - ForceFromGravity = 0

Try that, and see what you get.

Dorothy

Yes I'm using 10m/s*s for gravity

I truly understand now, thank you very much for your help I had trouble with this one since the day before Thanksgiving. I really appreciate the help.

sheebajenny said:
correction then I first found acceleration and then I used that to find normal force and then force of friction for the second problem one so I got normal force=325N and force of friction as 70.4N

Could we focus on one problem at a time

This normal force is much too low for an 80 kg weight on a 37 degree slope.

## 1. What is the normal force?

The normal force is the force that a surface exerts on an object that is in contact with it. It is perpendicular to the surface and acts in the opposite direction of the force applied by the object.

## 2. How is the normal force calculated?

The normal force can be calculated by using the formula FN = mgcosθ, where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle between the surface and the object.

## 3. What is the force of friction?

The force of friction is the force that opposes the motion of an object when it is in contact with a surface. It is caused by the roughness of the surfaces in contact and acts in the opposite direction of the motion of the object.

## 4. How is the force of friction calculated?

The force of friction can be calculated by using the formula Ff = μFN, where μ is the coefficient of friction and FN is the normal force.

## 5. What is the relationship between acceleration and the forces?

According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that the greater the force applied, the greater the acceleration, and the greater the mass, the smaller the acceleration.

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