Question

A wrench 30 cm long lies along the positive y-axis and grips 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 100 N*m of torque to the bolt.
I don't know what to do with <0,3,-4>?

Answer 1

well first we need dimensional homogeneity... torque is in newton meter... so change 30cm to meter first thing...

Answer 2

.3

Answer 3

then we know that torque is just a force times a distance... so we can set it uo like 100Nm=F(.3m)... from here we just divide by .3 to find that we need a force with a magnitude of 333.3 newtons in the <0,3,-4> direction... force is a vector and thus has a magnitude and a direction!

Answer 4

understand?

Answer 5

not really, where are you getting 333.3 newtons from?

Answer 6

First recall that torque = F x r

so since r ( the length of the wrench) is in the direction of the y-axis it can be expressed as a vector:

.3j

Since we know the direction of the force and not the magnitude we can say that F= 3aj-4k
Now you can solve for "a" using F x r= 100. then you can figure out the magnitude of F

Answer 7

*3aj-4ak *** sorry my mistake

Answer 8

ohhh ok, yeah I had no clue what torque was equal to. Thank you!! makes alot more sense now!

Answer 9

and its r X F not F x r lol opps another mistake

Answer 10

what would be the difference?

Answer 11

try it and see :)

Answer 12

ok. Thank you so much!

Answer 13

aww cant do 3 x 3 matrices on here

Answer 14

do you mean to do the cross product?

Answer 15

yes

Answer 16

i get , <.9a,0,0> for the vector and set it equal to 100?

Answer 17

Let me work it out.. havent done it yet



when you take the cross product of that matrix you will only get something in the i direction
\[i \left[\begin{matrix}.3 & 0\\ 3a& -4a\end{matrix}\right]\]

the determinant of that matrix leaves -1.2a i

the magnitude of that is just 1.2a