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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.
Figure 1. Angled force acting on a box that rests on a table.
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.
Figure 2. Components of a force F at angle θ on box with mass m. The component Fx is in the horizontal direction, and Fy is in the vertical direction.
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,Fx. We can apply Newton’s second law to the horizontal direction and write Fx in terms of F and θ.

Analyzing forces in the vertical direction

If the box stays on the table, the vertical component of F,Fy 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 Fy in terms of F and θ.

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.

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  • blobby green style avatar for user Sage Bauland
    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)
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    • hopper cool style avatar for user obiwan kenobi
      -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)
  • male robot johnny style avatar for user Sofia Prencipe
    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)
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  • blobby green style avatar for user zeljaoussi2026
    Why is the y-component of the force in the video

    (2 votes)
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  • male robot hal style avatar for user Araa$h
    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)
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  • starky tree style avatar for user robishen000
    What does F sub p stand for is it applied force?
    (1 vote)
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  • leaf blue style avatar for user Bruce Gabudao
    is a_y the same as F_g?
    (0 votes)
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    • area 52 yellow style avatar for user aviva
      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)
  • blobby green style avatar for user austingae
    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)
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  • duskpin sapling style avatar for user Violet
    This is too hard for me to understand :c,
    is there a simplified answer?
    (0 votes)
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    • blobby green style avatar for user JT
      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)
  • blobby green style avatar for user Anastasia  Akapn
    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)
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  • blobby green style avatar for user Jenna  Abu Ali
    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)
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