Question

Torque of a force F = 10(î– k) acting on point (1,0,2) about a point (1,1,0) is?

Answer 1

AnswEr :

From the Question,

• Force = 10(i - k) N

• Final Position = (1,1,0) = ( i + j ) m

• Initial Position = (1,0,2) = (i + 2k) m

We have to find the torque on the system

Firstly,

Displacement = Final Position - Initial Position

» Displacement = (i + j) - (i + 2k)

» Displacement = j - 2k

» r = ( j - 2k ) m

Now,

$\sf \tau \: = \vec{r} \: \times \vec{f} \\ \\ \longrightarrow \: \sf \: \tau = (0 \hat{i} + \hat{j} - 2 \hat{k}) \times (10 \hat{i} + 0\hat{j} - 10 \hat{k}) \\ \\ \longrightarrow \: \sf \: \tau = 10 \hat{i} \hat{j} - 10 \hat{i} \hat{k} - 20 \hat{i} \hat{j} \\ \\ \longrightarrow \: \sf \tau = -10 \hat{i} \hat{j} - 10 \hat{i} \hat{k} \\ \\ \longrightarrow \sf \tau = -10(k) - 10(-j) \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \tau = 10\hat{j} - 10 \hat{k} \: Nm}}$

Torque acting on the system is 10(j - k) Nm