Question

Sam applies a force at 30° to the horizontal to move a football tackling dummy 10 m horizontally.

He does 160 J of mechanical work. Determine the magnitude of the force by using vector dot product.

Answer 1

`F=(32√(3))/(3)\text{\:or\:} \approx 18.48\:\text{N}`

Explanation:

We can find the horizontal component of force he applied by using the formula for work:

`W=F\Delta x`

Solving for force, we have:

`160=F\cdot 10,\\F=16\:\text{N}`

However he applied the original force at an angle of
`30^(\circ)` to the horizontal. This force of 16 newtons is only the horizontal component of that force. To find the magnitude of the original force, we can use basic trigonometry:

`\cos 30^(\circ)=(16)/(x),\\x=(16)/(\cos 30^(\circ)),\\\\x=(16)/((√(3))/(2)),\\\\x=16\cdot(2)/(√(3))=(32)/(√(3))=\boxed{(32√(3))/(3)}`