Am i right on this? Newtons secocd law hw problem

  • Thread starter Thread starter mr.coon
  • Start date Start date
  • Tags Tags
    Law Newtons
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem related to Newton's second law, specifically involving forces acting on a boat. The problem includes calculating the boat's acceleration based on the forces applied in different directions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster questions how to determine the sign of forces when adding them, considering their directions on the xy-axis and the object's intended travel direction. Participants discuss their interpretations of positive and negative forces based on direction.

Discussion Status

Some participants have provided guidance on defining positive and negative forces, while others have shared their calculations and reasoning. Multiple interpretations of the problem setup are being explored, but there is no explicit consensus on the final answer.

Contextual Notes

Participants mention the context of the homework problem and the original poster's repeated attempts at the class, indicating a potential struggle with the material. There is an emphasis on understanding the application of Newton's laws rather than just arriving at a solution.

mr.coon
Messages
16
Reaction score
0
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 homework 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.
 
Physics news on Phys.org
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 ^^
 
thanks for the quick reply. i am taking this class for the third time and i am finally reaching out for help.

so sense 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
 
Yup , U R right ^^


We are here anytime . Feel Free to ask .

May God Help U In your Study ^^
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
4K
Replies
13
Views
3K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
6
Views
3K
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K