Please help with fiding the torque about an orgin and a point of a particle

[qwerty11]

Homework Statement

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

The problem is:

A particle is located at the 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?

Homework Equations

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.

 

[PhanthomJay]

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.

 

[qwerty11]

So it would be 10Nm at the orgin?

 

[PhanthomJay]

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.

 

[qwerty11]

Could you give me the equation you are referring to?

 

[PhanthomJay]

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).

 

[qwerty11]

A particle is located at the 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.

 

[qwerty11]

Bump.

 

[PhanthomJay]

A particle is located at the 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.

 

[qwerty11]

But isn't the 3j the vertical force and so since point Q is 2 integers above point P (y=6 vs y=4) shoudn't the vertical 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.

 

[qwerty11]

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

Thanks

 

[PhanthomJay]

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.