Question

The forces acting on a sailboat are 390n north and 180n east. if the boat (including crew) has a mass of 270 kg, what are the magnitude direction of the boat's acceleration

Answer 1

F=ma so a=F/m

ax=180/270=0.67m/s^2
ay=390/270=1.44m/s^2

Magnitude = sqrt((0.67^2)+(1.44^2))=1.59m/s^2

Direction- Tan(x)=0.67/1.44=0.47 Tan^-1(x)=25 degrees

Answer 2

Step-by-step explanation:

It is given that,

Force acting due north direction,
`F_N=390\ N`

Force acting due east direction,
`F_E=180\ N`

Mass of the boat, m = 270 kg

Let F is the magnitude of force acting on the sailboat. The resultant force acting on the sailboat is given by :

`F=√(F_N^2+F_E^2)`

`F=√(390^2+180^2)`

F = 429.53 N

Since, F = ma

`a=(F)/(m)`

`a=(429.53\ N)/(270\ kg)`

`a=1.59\ m/s^2`

Let
`\theta` is the direction of the boat's acceleration. It is given by :

`tan\theta=(F_N)/(F_E)`

`tan\theta=(390)/(180)`

`\theta=65.22^(\circ)`

Hence, this is the required solution.