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?

You would only substitute in the horizontal component of the force because the components of the force are independent from each other.

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?

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.

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?

-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!

is a_y the same as F_g?

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!

This is too hard for me to understand :c, is there a simplified answer?

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`

Main content

Course: High school physics > Unit 3

Lesson 6: Angled forces

Angled forces review

Google Classroom

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:

\begin{aligned} F_{x} & = F \cos θ \\ and \\ F_{y} & = F \sin θ \end{aligned}

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

θ

m a_{x} = F_{x} = F \cos θ

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

θ

\begin{aligned} m a_{y} & = F_{y} \\ 0 & = F \sin θ + F_{N} - F_{g} \end{aligned}

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?

Sort by:

Sage Bauland
Posted 4 years ago. Direct link to Sage Bauland's post “I was told to use -9.8 as...”
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?
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- obiwan kenobi
  Posted 4 years ago. Direct link to obiwan kenobi's post “-9.8 m/s^2 is not the for...”
  -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!
  Button navigates to signup page
  (11 votes)
Sofia Prencipe
Posted 5 years ago. Direct link to Sofia Prencipe's post “If you are trying to find...”
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?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- obiwan kenobi
  Posted 4 years ago. Direct link to obiwan kenobi's post “You would only substitute...”
  You would only substitute in the horizontal component of the force because the components of the force are independent from each other.
  Button navigates to signup page
  (3 votes)
zeljaoussi2026
Posted 2 months ago. Direct link to zeljaoussi2026's post “Why is the y-component of...”
Why is the y-component of the force in the video
�
�
�
�
�
Fsinθ?
Button navigates to signup pageComment on zeljaoussi2026's post “Why is the y-component of...”
(2 votes)
Answer
Araa$h
Posted 5 years ago. Direct link to Araa$h's post “This question rose to my ...”
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?
Button navigates to signup pageComment on Araa$h's post “This question rose to my ...”
(1 vote)
Answer
- brian
  Posted 4 years ago. Direct link to brian's post “There is no normal force ...”
  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.
  Comment on brian's post “There is no normal force ...”
  (3 votes)
robishen000
Posted 5 months ago. Direct link to robishen000's post “What does F sub p stand f...”
What does F sub p stand for is it applied force?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- masterYoda
  Posted 3 months ago. Direct link to masterYoda's post “F usually stands for Forc...”
  F usually stands for Force. Sub P would specify which force it is. Maybe theres a force that p ushes on something?
  Button navigates to signup page
  (1 vote)
Bruce Gabudao
Posted a year ago. Direct link to Bruce Gabudao's post “is a_y the same as F_g?”
is a_y the same as F_g?
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- aviva
  Posted 10 months ago. Direct link to aviva's post “Not necessarily. For obje...”
  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!
  Button navigates to signup page
  (2 votes)
austingae
Posted 5 years ago. Direct link to austingae's post “If a block is free fallin...”
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?
Button navigates to signup pageComment on austingae's post “If a block is free fallin...”
(1 vote)
Answer
- Sukhada
  Posted 8 months ago. Direct link to Sukhada's post “No, only 2 forces are act...”
  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.
  Button navigates to signup page
  (1 vote)
Anastasia Akapn
Posted 3 years ago. Direct link to Anastasia Akapn's post “How do you find the verti...”
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?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Tanisha
  Posted 3 years ago. Direct link to Tanisha's post “Knowing that F=ma, find t...”
  Knowing that F=ma, find the magnitude of Fsubp and then Fsubp sin θ.
  Button navigates to signup page
  (1 vote)
Jenna Abu Ali
Posted 5 years ago. Direct link to Jenna Abu Ali's post “A 45.0 kg box is pulled w...”
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.
Button navigates to signup pageComment on Jenna Abu Ali's post “A 45.0 kg box is pulled w...”
(0 votes)
Answer
Violet
Posted 9 months ago. Direct link to Violet's post “This is too hard for me t...”
This is too hard for me to understand :c,
is there a simplified answer?
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- JT
  Posted 8 months ago. Direct link to JT's post “Sure! Think the an angled...”
  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
  Button navigates to signup page
  (5 votes)