Question

Determine the moment of the couple for the system shown below. The forces shown lie in the x-y plane.

Answer 1

Given that,

Force = 800 lb

Distance = 3 ft

According to the figure,

The forces are,

$\vec{F_{1}}=F(-\sin\theta\i+\cos\theta\ j)$

$\vec{F_{2}}=F(\sin\theta\ i-\cos\theta\ j)$

Where, F = force

$\theta$ = angle between forces

The forces are acting at points

$\vec{r_{1}}=l\ j$

$\vec{r_{2}}=l\ j+\dfrac{3}{\sin\theta}\ j$

We need to calculate the moment of the couple

Using formula of net torque

$\tau=\vec{r_{1}}\times F_{1}+\vec{r_{2}}\times F_{2}$

Where, F = force

r = perpendicular distance

Put the value into the formula

$\tau=Fl\sin\theta\ k+F(l\sin\theta+\dfrac{3}{\sin\theta}\times\sin\theta)\ (-k)$

$\tau=-3F\ k$

$\tau=-3\times800$

$\tau=-2400\ k\ ft.lb$

Hence, The moment of the couple is -2400k ft.lb