What is the acceleration of the 1000 kg boat?

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Homework Help Overview

The discussion revolves around calculating the acceleration of a 1000 kg boat subjected to two forces: a 2000 N motor force and an 1800 N resistive force. Participants are analyzing the application of Newton's second law and vector components in this context.

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  • Mixed

Approaches and Questions Raised

  • One participant attempts to resolve the problem using vector components, while others question the necessity of this approach, suggesting a simpler method by directly considering the net force. There is also a discussion about the correct application of trigonometric functions in the calculations.

Discussion Status

Participants are exploring different methods for calculating the acceleration, with some providing guidance on simplifying the approach. There is no explicit consensus on the best method, but constructive feedback is being exchanged regarding the use of vectors and the importance of accuracy in calculations.

Contextual Notes

One participant notes that the original poster may be following a specific method taught in class, which could be contributing to the complexity of their approach. There is also a suggestion to draw a diagram to aid understanding.

atse1900
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A boat moves through the water with two forces acting on it. One is a 2000 N motor, and the o ther is an 1800 N restive force to the water. 1) What is the accelration of the 1000 kg boat?

Did I do this right?

a) (2000 cos 90 + 1800 cos 270 ) i + (2000 sin 90 + 1800 sin 270)j = 3800j N

sqrt(3800^2+0^2)= 3800 N.

a= 3800/1000 = 3.8 m/s^2
 
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The net force is 200 N forward. What are you doing with vectors? That's making this way more complicated than it really is.
 
atse1900 said:
A boat moves through the water with two forces acting on it. One is a 2000 N motor, and the o ther is an 1800 N restive force to the water. 1) What is the accelration of the 1000 kg boat?
Did I do this right?
a) (2000 cos 90 + 1800 cos 270 ) i + (2000 sin 90 + 1800 sin 270)j = 3800j N
sqrt(3800^2+0^2)= 3800 N.
a= 3800/1000 = 3.8 m/s^2
As durt pointed out you made this far more complicated than necessary. Also, the way you are trying to use vectors seems unnecessarily complicated, however it should have worked. Your mistake was that sin 270 = -1, not 1. Fix that and you'll get the right answer.

It looks like you are just using a cookbook method given to you by your teacher (well, that's the only reason i can think of to intentionally make it that complicated). It will always work (if you take the sines and cosines correctly) but in this case its just more work than necessary. In this case you have two opposing forces, since one is opposite direction change the sign. Since both are at the same angle, just add them directly. Much simpler.
 
Last edited:
Always draw a picture. It helps tremendously.
 

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