Question

I don't want the answer, just the steps to get there. Thanks for your help in advance.

A wrench 30 cm long lies along the positive y-axis and grips a bolt at the origin. A force is applied in the direction <0, 3, -4> at the end of the wrench. Find the magnitude of the force needed to supply 100N/m of torque to the bolt.

Answer 1

Post this in Physics

Answer 2

Its Math 231 (Motion in Space: Velocity and Acceleration)

Answer 3

not saying it aint math, but a post up in physics would be useful as well since its a physics application :)

Answer 4

Alright, I'll give it a try.

Answer 5

@amistre64
"force is applied in the direction <0, 3, -4> at the end of the wrench."

[considering only y-z plane]
this would mean that vector F passes through (3, -4) and (30, 0) right?

Answer 6

it would appear to me that the magnitude of the downward force, times the length of the wrench

Vaide, i believe so, yes

Answer 7

well, the vector need not pass thru 3,-4; but its direction from the end is such

Answer 8

Initial point (0i, 30j, 0k)
< 0, 3, -4> = Velocity? Acceleration?
If Acceleration:
F=ma -> F/m = a -> 100N/m = < 0, 3, -4>?
Then anti-derivative of < 0, 3, -4> to Velocity -> then take the magnitude of the Velocity?

Answer 9

so we can find the slope of F which will be equal to tan(theta). from this we can know theta.
so, T = F.s. sin(theta)

Answer 10

id have to read up on torque to be certain :)

Answer 11

@MCouch don't proceed that way. use the formula for torque.

Answer 12

torque = distance cross force

Answer 13

here's a brush-up:
torque = r x F

Answer 14

so, T = F.r.sin(theta)
we know T, r, theta. find F.

Answer 15

x 0 0 x = -120
y 30 3 y = 0
z 0 -4 z = 0

as is; you seem to have 120 somethings

Answer 16

to scale the force use 0s 3s and -4s is my idea

x 0 0s x = -120s
y 30 3s y = 0
z 0 -4s z = 0

where 120s = 100; or s= 100/120, 5/6

Answer 17

@amistre64 That makes more sense to me
@Vaidehi09 We have not gone over torque in that way yet...

Answer 18

the force direction creates a 3,4,5 right triangle; so the actual force might be 5s

Answer 19

oh? well, then use whichever way suits u!

Answer 20

i used this as a stepping stone

http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html

Answer 21

still not quite sure how to convert that to Nm

Answer 22

id imagine 30cm = .3 meters which would have to be used in the stuff <0,.3,0> instead

Answer 23

In my notes from a previous problem we did in class, we had 20N = 4kg x a we then converted it into a(t) = < 0, 0, 5> b/c it was stated the force was only in the positive z-axis

We then did the anti-derivative of a(t) to find v(t)

Answer 24

im not sure how to convert those to at and vt :/

Answer 25

Well thanks for the help thus far

Answer 26

one newton/meter is equal to the torque applied by one newton on a lever of one meter is what the wiki says ...

Answer 27

if we shorten the lever, we lessen the torque right?

in any event, im going with distance in meters crossed with the force vector :)

Answer 28

is there an answer key we can chk? cause i think i come up with about 416

Answer 29

No answer key :( and i'm trying to get in touch with a class mate.

And this is" MATH 231 Calculus of Several Variables (2) Analytic geometry in space; partial differentiation and applications. " If that helps

Answer 30

my thoughts are:

i define torque as the amount of pressure at the end of the lever;

that pressure is calculated as length * perpendicular force.

a NewtonMeter is a length of 1 meter, with a force of ! newton

30 cm = .3m ; the perp force is the "4" portion of the force vector; and the actual force of the vector is the hypotenuse of the 3,4,5 triangle it creates (the 5 part)

100 = .3 * 4s

100 = 1.2s

s = 1000/12

5s = 5000/12 = 416 2/3

not that its right or anything mind you :)

Answer 31

I will keep that in mind maybe it is right :)

But either way I'm going to take a break and hopefully my classmate will respond. I will also continue to dig through my textbook and notes for something that will give me heads or tails of this...

Answer 32

Thanks again.

Answer 33

good luck ;)

Answer 34

@amistre64 You were correct with 416.5.

You loosely follow F=ma or F/m =a
F/m = 100N/m and a = Starting point < 0, -.3, 0> cross directional vector < 0, 3, -4>
a = 83.3 and you have to do <0a, 3a, -4a> to get the vector and then take the magnitude of it.