Forces Acting at an Angle
The equations used to calculate the horizontal and vertical components of a force F acting at an angle θ measured from the positive x-axis are:

If the angle given is actually a reference angle, α, to the nearest x-axis instead of the directional angle θ (which is always measured counterclockwise from the positive x-axis), you must decide whether Fxand Fy are negative. Remember that forces are measured in newtons.

In the example shown above, both the x-component and the y-component of F are pointing in negative directions. Usually, the magnitude of the component is labeled without the use of a negative sign (-) since it is conveyed graphically by the direction of the component's vector. The following diagram is provided to assist you in remembering which functions are positive in which quadrants.

You can also recall these relationships by remembering that when the
• x-component of the reference angle points along the positive x-axis, cosine and its reciprocal function secant are positive
• y-component of the reference angle points along the positive y-axis, sine and its reciprocal function cosecant are positive.
Let's work a few examples.

 What are the magnitudes of the x- and y-components of the force vector, F = (10.0 nts, 75º)?

 What is the magnitude of the x-component of the force vector F = (10.0 nts, 105º)?

 What is the magnitude of the y-component of the force vector F = (10.0 nts, 255º)?

Problems involving "diagonal" forces

Let's begin by examinine a force F pulling up at an angle θ on one side of an object which is located on a rough horizontal plane. A freebody diagram of this situation is shown below.

Our first step will be to resolve, or break down, the diagonal force into its horizontal and vertical components. These components, Fx and Fy are shown in the next diagram.

Since the object is in vertical equilibrium, the relationship ΣFy = 0 will produce the equation

N = mg - F sin θ

If F has been pushing down at angle θ on the object instead of pulling up at angle θ,

then the value for the normal would change to

N = mg + F sin θ.

If the object is also in horizontal equilibrium, that is, it is either at rest or moving with a constant velocity towards the right, the relationship ΣFx = 0 yields the equation

f = F cos θ where f = µN.

If the object were accelerating to the right, the equation that could be used to calculate the object's horizontal acceleration would be

net Fx = ma
T cos θ - f = ma where f = µN.

Notice that the magnitude of the acceleration will be changed by the value of f since it is impacted by the value of the normal. Hence, it is easier to accelerate an object by pulling up on it (decreasing the normal force) than by pushing down on it (increasing the normal force).

Refer to the following information for the next two questions.

Suppose a 3-kg block is being pushed against a wall by a force F = 15 N acting at an angle of 30º to the horizontal.

 What is the normal force of the wall on the block?

 What is the magnitude of the friction present between the block and the wall?

Refer to the following information for the next five questions.

Using the system shown below, what would be the maximum mass, M, that can be supported by the vertical rope if the coefficient of static friction between the table and the 10-kg block is 0.4?

We will start our solution by examining freebody diagrams and the static equilibrium equations: ΣFx = 0 and ΣFy = 0. Since tensions are 3rd law forces we can find the tension on either side of a "rope" and it will be constant throughout the rope.

 T3 = Mg T2 = T1 cos 60 T1 sin 60 = T3 fs = µN N = mg f = T2
 What is the maximum, or critical, value for the static friction present between the 10-kg block and the table?

 What is the value of T2?

 What is the value of T1?

 What is the value of T3?

 What is the value of the suspended mass, M?

 Other common applications of forces applied at angles are objects moving along incline planes, simple pendulums, conical pendulums, and systems of bodies with knots and pulleys. Each of these topics are included in their own lessons which are hyperlinked in the resource lessons listed under Related Documents at the bottom of the page. Note that when working with "frictionless, massless" pulleys, the tensions on either side of the pulley are the same and should be labeled identically. NO COMPONENTS should be calculated for ropes as they go into or come out of frictionless, massless pulleys regardless of any angles present.