Calculate Torque of Force F Applied at r Around Origin

In summary: So, in summary, the problem is asking for the torque produced by a force applied at a point (4.0x, 7.0y) meters from the origin. This can be calculated using the formula Torque = r * F * sin theta, where r is the magnitude of the radius and F is the magnitude of the force, and theta is the smallest angle between the directions of r and F. Alternatively, the torque can also be found using the cross product formula, which involves finding the determinate of a matrix.
  • #1
uchicago2012
75
0

Homework Statement


The force F = (15x - 22y)N is applied at r = (4.0x + 7.0y) meters. What torque (about the origin) does this force produce?

Homework Equations


Torque = r * F * sin theta
r = magnitude of radius, F = magnitude of F, theta = smallest angle between the directions of r and F

The Attempt at a Solution


I was wondering what "about the origin" was referring to. Theoretically, shouldn't I be able to find the magnitudes of F and r with (ax2 + ay2)1/2 and the angles they make with the positive x-axis with (tan theta = ay/ax) and then figure out the smallest angle between them from that and calculate the torque?

Is it just saying the origin is the rotational axis so I don't have to worry about trying to translate the equations to some other rotational axis?
 
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  • #2
uchicago2012 said:
I was wondering what "about the origin" was referring to. Theoretically, shouldn't I be able to find the magnitudes of F and r with (ax2 + ay2)1/2 and the angles they make with the positive x-axis with (tan theta = ay/ax) and then figure out the smallest angle between them from that and calculate the torque?
Yes, that is a completely valid way to solve the problem. Find magnitudes of F and r, and the angle between them, then plug those into your formula. Then use the right hand rule to determine the direction of the resulting torque.

Or, alternately (which is what I would do because I think it's a little easier), to use the definition of torque, which is

[tex] \vec \tau = \vec r \times \vec F [/tex]

and recognize that the definition of the cross product is,

[tex]
\vec a \times \vec b =
\left|
\begin{array}{ccc}
\hat x & \hat y & \hat z \\
a_x & a_y & a_z \\
b_x & b_y & b_z
\end{array}
\right|
[/tex]

where the right side of the equation is the determinate of the matrix.

Either way should give you the correct answer.
Is it just saying the origin is the rotational axis so I don't have to worry about trying to translate the equations to some other rotational axis?
Essentially, yes. That's a good way to think of it. The length vector r has two points. One point at the origin and the other at (4.0, 7.0, 0.0). [And the force is applied at point (4.0, 7.0, 0.0).]
 

FAQ: Calculate Torque of Force F Applied at r Around Origin

What is torque?

Torque is a measure of the turning or twisting force on an object. It is calculated by multiplying the force applied to an object by its distance from the point of rotation.

How is torque calculated?

To calculate torque, you must know the magnitude of the applied force and the distance between the point of rotation and the line of action of the force. The formula for torque is T = r x F, where T is torque, r is the distance from the point of rotation, and F is the force applied.

What is the unit of measurement for torque?

The SI unit for torque is Newton-meters (Nm), but it can also be expressed in other units such as foot-pounds (ft-lb) or pound-feet (lb-ft).

Can torque be negative?

Yes, torque can be positive or negative depending on the direction of the applied force and the direction of rotation. A positive torque will cause an object to rotate in a counterclockwise direction, while a negative torque will cause it to rotate in a clockwise direction.

How does the distance affect torque?

The distance from the point of rotation to the line of action of the force is directly proportional to the torque. This means that as the distance increases, so does the torque. Therefore, a force applied at a greater distance will have a greater torque than the same force applied at a shorter distance.

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