Question

Two forces act simultaneously on a 20 kg box. If the forces are 40 N and 30 N, calculate the 3 possible accelerations of the box. (Forces in the same direction, forces opposite each other and forces perpendicular to each other .)

Answer 1

First case forces in the same direction

`F=30+40=70N`
`F=ma`

We isolate the a

`a=(F)/(m)`

Then we substitute the data

`a=(70)/(20)=3.5m/s^2\text{ }`

Second case

`F=40-30=10N`
`a=(10)/(20)=(1)/(2)=0.5m/s^2`

Third case

`F_x=40N`

`F_y=30-W`

The weight W is

`W=20\cdot9.8=196N`
`F_y=30-196=-166`
`R=\sqrt[]{40^2+(-166)^2}=170.75N`
`\theta=\tan ^(-1)(-166)/(40)=-76.45\text{\degree}`
`a=(F)/(m)=(170.75)/(20)=8.5375m/s^2\text{ at -76.45\degree}`