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### Course: Class 11 Physics (India) > Unit 9

Lesson 5: Normal force and contact force- Normal force and contact force
- Normal force in an elevator
- Breaking down forces for free body diagrams
- Free body diagram with angled forces: worked example
- Normal force (basic)
- Forces at an angle
- Angled forces review
- More on Normal force (shoe on floor)
- More on Normal force (shoe on wall)
- What is normal force?

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# Angled forces review

Review the key skills for angled forces, such as how to break down forces into the horizontal and vertical components.

## How to write force equations using components of an angled force

Sometimes forces are angled and do not point along the coordinate axes. Let's analyze the specific example shown in Figure 1.

An angled force can be broken down to horizontal and vertical components (see Figure 2 below). This allows us to apply Newton’s second law to the forces in the horizontal and vertical directions separately.

The components of the applied force $F$ are:

### Analyzing forces in the horizontal direction

If our box in Figure 1 experiences no friction, the only force acting horizontally is the horizontal component of $F,{F}_{x}$ . We can apply Newton’s second law to the horizontal direction and write ${F}_{x}$ in terms of $F$ and $\theta $ .

### Analyzing forces in the vertical direction

If the box stays on the table, the vertical component of $F,{F}_{y}$ acts vertically along with weight down and normal force up to produce zero vertical acceleration. We can apply Newton’s second law to the vertical direction and write ${F}_{y}$ in terms of $F$ and $\theta $ .

## Learn more

To see a worked example involving a force at an angle, see our video about tension forces on a box.

To check your understanding and work toward mastering these concepts, check out the exercise on forces at an angle.

## Want to join the conversation?

- I was told to use -9.8 as the force of gravity but these videos say that I should use the opposite of the normal force or the weight.

Also how do I factor in mass when calculating these problems?(0 votes)- -9.8 m/s^2 is not the force of gravity, it is the free fall acceleration due to gravity on Earth. According to Newton's second law, F = ma. Which means that Weight = mass * gravitational acceleration = m * 9.8 m/s^2. Now, you put in the mass of the object in kilograms to get the weight of the object in Newtons. Hope this helps!(11 votes)

- If you are trying to find the acceleration of an object and the force is being applied at an angle do you substitute the total force for f in the a=f.m formuala or only the horizontal force that you find using cosine?(2 votes)
- You would only substitute in the horizontal component of the force because the components of the force are independent from each other.(3 votes)

- Why is the y-component of the force in the video

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Fsinθ?(2 votes) - This question rose to my mind because of an answer I had to someone else's: If an object is free-falling isn't there technically normal force with all of the air particles, or is the normal force of these air particles considered air resistance?(1 vote)
- There is no normal force acting upon an object if it is in the air. Also, there is no air resistance in free fall, the only force acting on the object has to be gravitational force.(3 votes)

- What does F sub p stand for is it applied force?(1 vote)
- F usually stands for Force. Sub P would specify which force it is. Maybe theres a force that
**p**ushes on something?(1 vote)

- is a_y the same as F_g?(0 votes)
- Not necessarily. For objects in freefall, the answer is yes, because gravity (F_g) is the only force acting upon the object. However, if there is a vertical or angled force other than gravity acting upon the object, a_y will not equal F_g.

To visualize this- think of a cup sitting on a table. F_g is acting upon the object, but so is the normal force that the table exerts on the object. Since the normal force (F_n) is equal to F_g, the block has no vertical acceleration, so a_y=0. Hope this helps!(2 votes)

- If a block is free falling at a certain height, are the two forces acting on the block the gravity and the air resistance? And there's no normal force because the block is accelerating downwards, right?(1 vote)
- No, only 2 forces are acting, since there is motion only in the y component. If the body is free-falling, then yes, it has only air resistance and gravity.(1 vote)

- This is too hard for me to understand :c,

is there a simplified answer?(0 votes)- Sure! Think the an angled force as a right triangle, shown above in Figure 2. In trigonometry, you learned these two formulas:
`sinθ = Opposite/Hypotenuse`

cosθ = Adjacent/Hypotenuse

Where Opposite is our vertical upward force, and our Adjacent is our horizontal rightward force.

Looking back at the triangle, we can see the Hypotenuse is our Force variable F, so we can replace Hypotenuse with F.`sinθ = Opposite/F`

cosθ = Adjacent/F

Multiply both sides of the equation by F to solve for the Opposite or Adjacent:`Opposite = Fsinθ`

Adjacent = Fcosθ

As you can see, our upwards force has a value of`Fsinθ`

and our rightwards force has a value of`Fcosθ`

. Now we just have to group the two forces with their respective directions.

For horizontal forces, since there is no friction, then there is no other horizontal forces, so our horizontal force would just be`Fcosθ`

.

For vertical forces, we have the force of gravity and normal force, so we can just add them together with our newly found upwards force.`Fsinθ + Normal Force + Force of Gravity`

Fsinθ + FN + (-Fg)

Fsinθ + FN - Fg

All Horizontal Forces:`Fcosθ`

All Vertical Forces:`Fsinθ + FN - Fg`

(6 votes)

- How do you find the vertical component of a pull at a 22-degree angle when only given the mass of the object, the μ, and the acceleration?(1 vote)
- Knowing that F=ma, find the magnitude of Fsubp and then Fsubp sin θ.(1 vote)

- A 45.0 kg box is pulled with a force of 205 N by a rope held at an angle of 46.5 degrees to the horizental. The velcoity of the box increases from 1.00 m/s to 1.50 m/s in 2.50 s. Calculate the following

a) The net force acting horizontally on the box.

b) The frictional force acting on the box.

c) The horizontal component of the applied force.

d) The coefficent of Kinetic friction between the box and the floor.

If you can help please do so, Thank you.(0 votes)