A force of F = 3.00 i + 2.00 j N is applied to an object that is pivoted about a fixed axis aligned along the?

Question

A force of F = 3.00 i + 2.00 j N is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis.
(a) If the force is applied at the point r = (4.00 i + 3.00 j + 0 k) m, find the magnitude of the net torque about the z axis.
N·m
(b) What is the direction of the torque vector, ?
negative i
positive i
positive j
negative j
positive k
negative k

Answer 1

torque T = r cross F
T = (4.00 i + 3.00 j + 0 k) X (3.00 i + 2.00 j)
cross multiplying vectorilly
T = 12*(iXi)+ 9*(jXi)+0 + 8*(iXj)+6*(jXj)+0
T = 12*0+ 9*(- k) + 8*(k)+6*0
T = 9*(- k) + 8*(k) = - k
===========
net torque |T| = 1 N-m
direction >> negative k

Answer 2

T = r x F;

do the vector cross product. Use the matrix form of the cross product since you have vector components; see http://en.wikipedia.org/wiki/Vector_cross_product

Use the right-hand rule to determine the direction of T: r turns to F, and the T will be in the direction that a right-hand screw will turn if rotated in that manner.