Dynamics: Forces at 90 and 45 degrees

In summary, the acceleration of the particle when subjected to the given forces is 7.1 m/s^2 at 90° and 9.2 m/s^2 at 45°, with the smaller angle resulting in a greater acceleration. The equation F=m.a can be used to calculate the acceleration, and in the alpha=90° case, Pythagoras' theorem can be applied. In the alpha=45° case, a free-body diagram should be used to consider the components of the forces.
  • #1
sv_heavymetal
2
0

Homework Statement


Hi, I have a system of forces which is showed in the following picture:

http://imageshack.us/photo/my-images/197/fuerzas.JPG/

The exercise says: A force F produces an acceleration of 5 m/s^2 when it acts on a particle of mass m. Calculate the acceleration of the particle when subjected to the forces as shown in the image, for the following cases: alpha: 90, alpha: 45.

As you can see, I did not give the values ​​of both F or m.

Results: The acceleration must be 7.1 m/s^2 to 90 ° and 9.2 m/s^2 to 45 °, that is, the smaller the angle the bigger the acceleration.

Homework Equations



F= m.a

The Attempt at a Solution

I think I have to use Pitagoras, using the 2 Fs as the legs and the resultant force will be the hypotenuse?

Thanks.
 
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  • #2
Welcome to PF!
sv_heavymetal said:

Homework Statement


Hi, I have a system of forces which is showed in the following picture:

http://imageshack.us/photo/my-images/197/fuerzas.JPG/

The exercise says: A force F produces an acceleration of 5 m/s^2 when it acts on a particle of mass m. Calculate the acceleration of the particle when subjected to the forces as shown in the image, for the following cases: alpha: 90, alpha: 45.

As you can see, I did not give the values ​​of both F or m.

Results: The acceleration must be 7.1 m/s^2 to 90 ° and 9.2 m/s^2 to 45 °, that is, the smaller the angle the bigger the acceleration.


Homework Equations



F= m.a

The Attempt at a Solution




I think I have to use Pitagoras, using the 2 Fs as the legs and the resultant force will be the hypotenuse?

Thanks.
Pythagoras only works when the two Fs are at 90°, since it only applies to a right triangle. So yes, you can use that for the alpha=90° case.

When alpha=45°, you should draw yourself a free-body diagram, and look at components of the individual forces.
 

FAQ: Dynamics: Forces at 90 and 45 degrees

What is the difference between a force at 90 degrees and a force at 45 degrees?

A force at 90 degrees is perpendicular to the surface it is acting on, while a force at 45 degrees is at an angle to the surface. This means that a force at 90 degrees will result in more work being done, while a force at 45 degrees will have a component of the force that is parallel to the surface.

How do forces at 90 degrees and 45 degrees affect an object's motion?

Forces at 90 degrees will result in a change in an object's direction, while forces at 45 degrees will result in both a change in direction and a change in speed. This is because forces at 45 degrees have both a perpendicular and parallel component, while forces at 90 degrees only have a perpendicular component.

Can forces at 90 and 45 degrees cancel each other out?

Yes, forces at 90 and 45 degrees can cancel each other out if they are equal in magnitude but opposite in direction. This is known as balanced forces. If the forces are not equal, they will result in a net force that causes an object to accelerate in the direction of the larger force.

What is the formula for calculating the net force of two forces at 90 and 45 degrees?

The formula for calculating the net force of two forces at 90 and 45 degrees is Fnet = √(F1² + F2² + 2F1F2cosθ), where F1 and F2 are the magnitudes of the forces and θ is the angle between them.

How do you determine the direction of the net force of two forces at 90 and 45 degrees?

The direction of the net force can be determined by using the tangent function. The angle of the net force will be tan^-1(F2sinθ / (F1 + F2cosθ)). This will give you the angle in relation to the larger force.

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