• qwerty11
In summary, the torque about the origin is 21 N-m, clockwise, and the torque about the point (0,6) is 15 N-m, counterclockwise.
qwerty11

Homework Statement

I just can't seem to get a handle on this.

The problem is:

A particle is located at teh vector position >r=(i+4j)m, and the force acting on it is >F=(6i+3j)N. What is the torque about the origin and the point having coordinates (0,6)m?

t=r x f

The Attempt at a Solution

((1)(3))-((3)(5)) = -14kNm at origin.

I believe this is wrong and have no idea how to do it at coordinate (o,6)m.

Any help would be greatly appreciated.

use torque = force times perpendicular distance. Draw a quick sketch on a graph. The force has an x component of 6 N, and a y component of 3 N. It is located at (1,4). Sum torques of each, and watch your plus/minus (cw/ccw) signs.

So it would be 10Nm at the orgin?

qwerty11 said:
So it would be 10Nm at the orgin?
How'd you get that number? Show your calculation. The torque of a force about a point is equal to the product of the force times the perpendicular distance from the line of action of that force to the origin. Clockwise is plus and Counterclockwise is minus, or vice versa. Algebraically add up the torques from the Fx and Fy components of the given force.

Could you give me the equation you are referring to?

qwerty11 said:
Could you give me the equation you are referring to?
Well heck Torque about origin is F_x(y) - F_y(x) = 6(4) - 3(1) = 21 N-m clockwise . Now you try it for the torque about the point (0,6).

A particle is located at teh vector position >r=(i+4j)m, and the force acting on it is >F=(6i+3j)N. What is the torque about the origin and the point having coordinates (0,6)m?

For the point (0,6) I get (10)6-3(1) = 57 N-m clockwise. I got the the 10 by doing 4j+6.

Bump.

qwerty11 said:
A particle is located at teh vector position >r=(i+4j)m, and the force acting on it is >F=(6i+3j)N. What is the torque about the origin and the point having coordinates (0,6)m?

For the point (0,6) I get (10)6-3(1) = 57 N-m clockwise. I got the the 10 by doing 4j+6.
Vectors are tough to work with, and i can see you are having a problem with them. The vector position of the particle, given as (1)i + 4j m, means that the particle is located at the coordinate x = 1 and y = 4. Plot it on a graph and mark it P. Now the force acting on that particle is given as 6i + 3j N, which means that the force has an x component of 6 N acting horizontally pointing toward the right, and it has a y component of 3 N acting vertically pointing up. So indicate those 2 forces on your graph, originating at point P. The 'line of action' of the 6 N force is a horizontal line passing thru point P; the 'line of action' of the 3 N force is a vertical line passing thru point P. Now for part a, you wanted the torque of these forces about the origin, O (the point at x=0, y=0). Recall that the torque of a force about a point is the force times the perpendicular distance from its line of action to the point. So the torque from the horizontal 6 N force is 6(4) =24 N-m, clockwise, and the torque from the vertical 3 N force is (3)(1) = 3 N-m, counterclockwise; thus, again, the total torque about O is 24 - 3 = 21 N-m, clockwise.

Now for part b, you want the torque of those forces about the point where (x=0, y=6). Plot that point and mark it Q. The forces still pass through P. Now using the same approach for determining torque, the torque about Q from the horizontal 6 N force is (6)(2) = 12 N-m, counterclockwise. Now what is the torque about point Q from the 3N vertical force? Once you find it, add 'em up to get the total torque about Q.
Of course, if you don't like this method, you can find the torque by the vector cross product r x F, which to me is a lot more difficult approach.

But isn't the 3j the verticle force and so since point Q is 2 integers above point P (y=6 vs y=4) shoudn't the verticle force be 3(2)=6 N'm counterclockwise and the horizontal force be -1(6) counter clockwise since Q is behind P?

Thanks for working with me though. I am really wanting to understand this.

I figured out how to use the cross product and got the answer. I am still interested to learn your method however.

Thanks

qwerty11 said:
I figured out how to use the cross product and got the answer. I am still interested to learn your method however.

Thanks
Ii suppose if you like the cross product method, go for it. I haven't used it since my college days, as an engineer, I like basic math and simple algebra, but as you wish. I hope you got 15N-m counterclockwise for part b. You probably just got a number, positive or negative, which you must interpret as cw or ccw. And i hope you got the right position vector between P and Q, or else you're dead in the water. By the alternate method, it's force times perpendicular distance for torque calculations. The 6N force is 2 meters away from Q, where the 2 m is the perpendicular distance between the horizontal line of action of the 6 N force and the point Q, forget if the force is below Q or wherever, it's the cw or ccw rotation of the torque that matters. The perpendicular distance between the vertical line of action of the 3 N vertical and point Q is 1 m, so its torque is (3)(1) = 3 N-m, also ccw, so the total torque about Q is 15 N-m, ccw.

What is torque?

Torque is a measure of the rotational force applied to an object.

How is torque calculated?

Torque is calculated by multiplying the force applied to an object by the distance from the point of rotation to the point where the force is applied.

What is the origin in terms of torque?

The origin in terms of torque is the fixed point or axis around which an object rotates.

What is the point of a particle in relation to torque?

The point of a particle in relation to torque is the specific location on the object where the force is applied.

How is torque used in physics and engineering?

Torque is used in physics and engineering to understand and predict the rotational motion of objects, as well as to design and analyze machines and structures that involve rotational movement.

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