Calculating useful force or force at an angle

In summary, the horizontal and vertical components of a force applied at an angle of 45° are equal in magnitude, but they are not half of the original force and do not necessarily add up to the original force in the same direction.
  • #1
richard9678
93
7

Homework Statement



I don't understand my textbook.

Homework Equations



Fh = Fa Cos θ

Where Fa is some pulling force applied, in-between horizontal and vertical.

Where Fh is force in the horizontal.

The Attempt at a Solution



Intuitively, if the angle of the force applied is at 45°, then the force in the horizontal and vertical is the same, and half of Fa.

Cos θ for 45° is 0.707. I was expecting 0.5.

What am I getting wrong? Thanks.
 
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  • #2
richard9678 said:
Intuitively, if the angle of the force applied is at 45°, then the force in the horizontal and vertical is the same, and half of Fa.
You are correct that the horizontal and vertical components must be the same, but that doesn't mean that they are each half the total force. Realize that they are perpendicular. Imagine a 45°-45° right triangle with sides of 1 unit length. What's the hypotenuse equal?
 
  • #3
richard9678 said:

Homework Statement



I don't understand my textbook.

Homework Equations



Fh = Fa Cos θ

Where Fa is some pulling force applied, in-between horizontal and vertical.

Where Fh is force in the horizontal.

The Attempt at a Solution



Intuitively, if the angle of the force applied is at 45°, then the force in the horizontal and vertical is the same, and half of Fa.

Cos θ for 45° is 0.707. I was expecting 0.5.

What am I getting wrong? Thanks.

Force components sum the way the legs of a right triangle "sum" to find the hypotenuse.

As you noted, if the angle is 45° then cos(θ) is 1/√2. But sin(θ) is also 1/√2, so the vertically directed force component is indeed equal to the horizontally directed component. But neither one is half of the net force!

For a right triangle with side lengths A and B, the hypotenuse C is given by the relationship:

C2 = A2 + B2

Force components add in the same fashion.

EDIT: Doc Al beat me to it!
 
  • #4
richard9678 said:
... if the angle of the force applied is at 45°, then the force in the horizontal and vertical is the same, and half of Fa...

Yes, at that angle, the two components of the original force have the same magnitude.

BUT since these two components are not in the same direction the magnitude of each of them is not half the magnitude of the original force.

Two forces, each of magnitude F/2, can only give a resultant of magnitude F if these two forces are in the same direction.
 
  • #5


Dear student,

Thank you for reaching out for clarification. It seems like you have a good intuition about calculating forces at an angle. Your understanding that the horizontal force is equal to half of the applied force at a 45° angle is correct. However, the equation you provided is for calculating the horizontal component of a force at any angle, not just at 45°.

The equation Fh = Fa Cos θ is derived from the concept of vector components. When a force is applied at an angle, it can be broken down into two components: the horizontal component and the vertical component. The horizontal component is the force acting in the horizontal direction, and the vertical component is the force acting in the vertical direction. The angle θ represents the angle between the applied force (Fa) and the horizontal direction.

To calculate the horizontal component of a force, we use the cosine function because it gives us the ratio of the adjacent side (horizontal component) to the hypotenuse (applied force). When θ is 45°, the cosine of 45° is indeed 0.707, which means that the horizontal component is equal to 0.707 times the applied force. However, for other angles, the cosine value will be different, and the horizontal component will also be different.

I hope this explanation helps you understand the equation better. If you have any further questions, please do not hesitate to ask. Keep up the good work in your studies!

Best regards,
 

What is useful force?

Useful force is the force that is actually doing the work in a given situation. It is the component of the applied force that is acting in the direction of motion or causing a change in motion.

How do you calculate useful force?

To calculate useful force, you need to know the applied force and the angle at which it is being applied. You can then use trigonometry to find the component of the applied force that is acting in the direction of motion or causing a change in motion.

What is force at an angle?

Force at an angle is when a force is applied in a direction that is not parallel to the direction of motion or the surface it is acting on. This results in the force being broken into two components: one that is parallel to the surface and one that is perpendicular to the surface.

How do you calculate force at an angle?

To calculate force at an angle, you need to know the magnitude of the force, the angle at which it is being applied, and the surface it is acting on. You can then use trigonometry to find the parallel and perpendicular components of the force.

Why is it important to calculate useful force at an angle?

Calculating useful force at an angle is important because it allows us to understand the amount of force that is actually being applied in a given situation. It also helps us determine the direction and magnitude of the force, which is crucial in many real-world applications such as engineering and physics.

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