am i right on this? newtons secocd law hw problem

[mr.coon]

this is from ch 4 of cutnell & johnson physics

when you add forces together such as the ones in the problem below, how do you know which ones are positive and which ones are negative? does it have to do with their relation on the xy axis? or does it have to do with the direction the object is trying to travel?

hw problem:

A boat has a mass of 6800 kg. Its engines generate a drive force of 3300 N, due west, while the wind exerts a force of 940 N, due east, and the water exerts a resistive force of 1200 N due east. What is the magnitude and direction of the boat's acceleration?

this is my attempt at the hw problem:

sum of the forces = -1610 then i divide that by the mass of the ship giving me -.237m/s^2 W as the a and direction of the ship.

[Fazza3_uae]

Hi mr.coon ^^

& Welcome to the PF

Let positive be west and negative be east since they are opposite directions.

The forces are 3300 , -940 , -1200

The sum of the forces : 3300-940-1200= 1160 N

Newtons law F=ma is used

F = 1160 N
m = 6800 Kg
a = Find It ^^

[mr.coon]

thanks for the quick reply. i am taking this class for the third time and i am finally reaching out for help.

so sence the ship is traveling west i set the opposing forces as negative:

net F = 3300N - 490N - 1200N = 1160N

then F= ma so a= F/m = 1160/6800 = 0.171 m/s squared due west

final answer = 0.171 m/s squared due west

[Fazza3_uae]

Yup , U R right ^^


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