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

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

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Muscardinus Muscardinus · Answer 2 · 2019-11-26T06:33:36+01:00

Explanation:

It is given that,

Force acting due north direction, [tex]F_N=390\ N[/tex]

Force acting due east direction, [tex]F_E=180\ N[/tex]

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 :

[tex]F=\sqrt{F_N^2+F_E^2}[/tex]

[tex]F=\sqrt{390^2+180^2}[/tex]

F = 429.53 N

Since, F = ma

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{429.53\ N}{270\ kg}[/tex]

[tex]a=1.59\ m/s^2[/tex]

Let [tex]\theta[/tex] is the direction of the boat's acceleration. It is given by :

[tex]tan\theta=\dfrac{F_N}{F_E}[/tex]

[tex]tan\theta=\dfrac{390}{180}[/tex]

[tex]\theta=65.22^{\circ}[/tex]

Hence, this is the required solution.