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}`