Question

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector vector r = (2.00 m)i hat - (3.00 m)j + (2.00 m)k, the force is vector F = Fxi hat + (7.00 N)j - (5.10 N)k and the corresponding torque about the origin is τ = (1.30 N·m)i hat + (-0.80 N·m)j + (-2.50 N·m)k. Determine Fx.

Answer 1

Answer:

$F_x=5.5\ N$

Explanation:

Given that,

Torque, $\tau=(1.3i-0.8j-2.5k)\ N-m$

Force, $F=(F_xi+7j-5.1k)\ N$

Position vector, $r=(2i-3j+2k)\ m$

Torque acting on the particle is given by :

$\tau=F\times r$

$(1.3i-0.8j-2.5k)=(F_xi+7j-5.1k)\times (2i-3j+2k)$

The coefficient of i should be same. So,

$(1.3i-0.8j-2.5k)=-(F_xi+7j-5.1k)\times (2i-3j+2k)$

$(1.3i-0.8j-2.5k)=\begin{pmatrix}1.3&10.2+2x&3x+14\end{pmatrix}$  

On comparing both sides, we get :

$F_x=5.5\ N$

So, the x component of force is 5.5 N. Hence, this is the required solution.