# How do i calculate the necessary force on a hill

• Mathijsgri
In summary, the problem involves a box on a 10 degree hill with a mass of 12,000 and a coefficient of friction of 0.08. The force needed to make the box move 20 meters is 30 kN, assuming the 80 degree angle refers to the angle of the hill. It is unclear where the information about wanting to achieve a speed of 1 m/s in 4 seconds and the force needed for constant speed comes from. More information, such as a free body diagram, is needed to properly solve the problem.
Mathijsgri

## Homework Statement

A box is on a 10 degree hill what is the force i have to pull to make the box move? and what force do i need to make the box move 20 meters

mass of the box: 12.000
Coefficient of friction: 0.08
speed 1 m/s

## Homework Equations

Ffriction= μ * Fnormal

F= m*a[/B]

## The Attempt at a Solution

Fnormal = Fgravtity* cos 80

Fn = (12,000* 10) * sin 80
Fn= 118177
Fn =118kN

Ffriction= μ * Fnormal

Ffriction= 0.08* 118
Ff= 9,45 kN

The total force to make the box move = 9,45 +120 * cos 80 = 30 kN

Is this correct?

F = m * a
I want to be at 1 m/s in 4s

F = 10.000 * 0,25
F= 2500N.

Do I have to pull 30kN to make the box move and then pull 2,5 kN or do i have to pull 32,5kN?

And when I achieve the 1 m/s what is the necessary force to make it a constante speed?

Last edited:
Special characters like "μ" are available in the menu that you can invoke by clicking on the ##\Sigma## icon in the edit window's top bar:

so no need to try to insert them as images from other sites. Such images often appear as broken links.

Mathijsgri said:

## The Attempt at a Solution

Fnormal = Fgravtity* cos 80

Fn = (12,000* 10) * sin 80
Fn= 118177
Fn =118kN
Where does the 80° angle come from? You didn't provide a drawing (or free body diagram) for the scenario that identifies the angles, so this introduction of 80° after saying that the hill is a "10 degree hill" is puzzling. Which angle is it?

Also, What is the difference between Fnormal and Fn above? One uses the cosine of 80° and the other the sine.

Ffriction= 0.08* 118
Ff= 9,45 kN

The total force to make the box move = 9,45 +120 * cos 80 = 30 kN

Is this correct?
That will depend on sorting out what the 80° angle is. You should draw and post a free body diagram.
F = m * a
I want to be at 1 m/s in 4s
Where did this come from? It wasn't in the problem statement. You shouldn't introduce new information during the solution attempt. Helpers need to see the complete problem statement ahead of the any solution attempt so that they can verify your approach and steps.
F = 10.000 * 0,25
F= 2500N.

Do I have to pull 30kN to make the box move and then pull 2,5 kN or do i have to pull 32,5kN?
What's the net force acting on the box in each case? What does Newton's 2nd law tell you about the motion of the box?
And when I achieve the 1 m/s what is the necessary force to make it a constante speed?
What are the conditions for constant speed? Again, look to Newton's 2nd law.

## 1. What is the formula for calculating the necessary force on a hill?

The formula for calculating the necessary force on a hill is F = mg sinθ, where F is the necessary force, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the hill.

## 2. How do I determine the mass of the object in the calculation?

You can determine the mass of the object by using a scale or by looking up the mass of the object in a reference book or online. The mass should be measured in kilograms (kg).

## 3. What is the value of acceleration due to gravity?

The average value of acceleration due to gravity on Earth is 9.8 m/s². However, this value may vary slightly depending on location and altitude.

## 4. Do I need to consider the weight of the object in the calculation?

Yes, the weight of the object is a necessary component in the calculation as it affects the amount of force needed to move the object up or down the hill. Weight is calculated by multiplying the mass of the object by the acceleration due to gravity (W = mg).

## 5. Can I use this formula for any hill or incline?

Yes, this formula can be used for any hill or incline as long as the angle of the hill is known. However, it is important to note that the formula assumes a frictionless surface and may not be completely accurate in real-life scenarios.

Replies
188
Views
8K
Replies
2
Views
502
Replies
2
Views
1K
Replies
9
Views
12K
Replies
5
Views
16K
Replies
20
Views
1K
Replies
8
Views
1K
Replies
1
Views
2K
Replies
21
Views
2K
Replies
12
Views
6K