Question

Three forces act on an object. Two of the forces are at an angle of 95◦ to each other and have magnitudes 35 N and 7 N. The third force is perpendicular to the plane of these two forces and has magnitude 9 N. Calculate the magnitude of the force that would exactly counterbalance these three forces. Round your answer to one decimal place.

Answer 1

We want to find $\vec F_4$ such that the object needs is in equilibrium:

$\vec F_1+\vec F_2+\vec F_3+\vec F_4=\vec0$

We're told that $F_1=35\,\mathrm N$, $F_2=7\,\mathrm N$, and $F_3=9\,\mathrm N$. We also know the angle between $\vec F_1$ and $\vec F_2$ is 95º, which means

$\vec F_1\cdot\vec F_2=F_1F_2\cos95^\circ=245\cos95^\circ$

$\vec F_3$ is perpendicular to both $\vec F_1$ and $\vec F_2$, so $\vec F_1\cdot\vec F_3=\vec F_2\cdot\vec F_3=0$.

If we take the dot product of $\vec F_1$ with the sum of all four vectors, we get

$\vec F_1\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0$

$\vec F_1\cdot\vec F_1+\vec F_1\cdot\vec F_2+\vec F_1\cdot\vec F_3+\vec F_1\cdot\vec F_4=0$

${F_1}^2+\vec F_1\cdot\vec F_2+0+\vec F_1\cdot\vec F_4=0$

$\implies\vec F_1\cdot\vec F_4=-\left({F_1}^2+\vec F_1\cdot\vec F_2\right)$

We can do the same thing with $\vec F_2$ and $\vec F_3$:

$\vec F_2\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0$

$\implies\vec F_2\cdot\vec F_4=-\left(\vec F_1\cdot\vec F_2+{F_2}^2\right)$

$\vec F_3\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0$

$\implies\vec F_3\cdot\vec F_4=-{F_3}^2$

Finally, if we do this with $\vec F_4$, we get

$\vec F_4\cdot(\vec F_1+\vec F_2+\vec F_3+\vec F_4)=0$

$\implies{F_4}^2=-\left(\vec F_1\cdot\vec F_4+\vec F_2\cdot\vec F_4+\vec F_3\cdot\vec F_4\right)$

$\implies{F_4}^2=-\left(-\left({F_1}^2+\vec F_1\cdot\vec F_2\right)-\left(\vec F_1\cdot\vec F_2+{F_2}^2\right)-{F_3}^2\right)$

$\implies F_4=\sqrt{{F_1}^2+{F_2}^2+{F_3}^2+2(\vec F_1\cdot\vec F_2)}$

$\implies\boxed{F_4\approx36.2\,\mathrm N}$