Question

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,what is the resulting torque on the pebble about (a) the origin and(b) the point (2.0m, 0,-3.0m)?a) dot product of F*r=(-3.0N)(-2.0m)=6.0Nmb) sqrt( 2.0^2+0^2+3.0^2) = sqrt(13)I've tried but not sure.Please help me.Thank you.

Answer 1

Answer with Explanation:

We are given that

Force acts on a pebble=$2\hat{i}-3\hat{k}$ N

Position vector=$r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque=$r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque=$-k-1.5i-4j$N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b.$r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$N-m