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- Thread starter Fatima Hasan
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- #1

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- #2

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Also, be advised that simply posting images of your work and of the problem is not compatible with the Homework guidelines of Physics Forums. Please type out the problem statement and your solution attempt in the future.

- #3

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thanks,

Also, be advised that simply posting images of your work and of the problem is not compatible with the Homework guidelines of Physics Forums. Please type out the problem statement and your solution attempt in the future.

but question is that my answer is not an exact of choice C. so is my answer right?

sorry but my skills are not good to write math formula

- #4

TSny

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I believe your final answer is correct. But it looks like you might have gotten lucky by having two compensating errors. Think about the direction of the friction force when the applied force is a minimum.

- #5

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It is close enough to one of the answers to suspect that you (or the authors) have a rounding error somewhere.but question is that my answer is not an exact of choice C. so is my answer right?

Really the formulas here are so simple that you should have no problem typing them out. If you click the Σ symbol you also have a choice of different useful symbols for writing maths. You do not need to use LaTeX for formulas that are this simple.sorry but my skills are not good to write math formula

By writing ##F = mg \sin(\theta) + F_s##, you are essentially writing "forces to the right = forces to the left". It is kind of artificially taking care of the signs by specifying all forces as positive and then placing them on the appropriate side of the equation. If I did not enter it wrong in Matlab, 16.9 N is the correct rounding of the answer, not 16.8 N.

I believe your final answer is correct. But it looks like you might have gotten lucky by having two compensating errors. Think about the direction of the friction force when the applied force is a minimum.

- #6

TSny

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Some instructors like to use 37thanks,

but question is that my answer is not an exact of choice C. so is my answer right?

So, the author of this problem might have intended you to use 0.80 and 0.60 for the cosine and sine. See if that gives you an answer of 16.8.

Edit: When I was a student ages ago, we didn't have calculators. So it was more common in those days to use these approximations when working with a 37

- #7

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$$

F = [60~\mbox{N}] \sin(\theta) - 0.4\cdot [60~\mbox{N}] \cos(\theta).

$$

In comparison to the approximation mentioned by @TSny, ##\sin(37^\circ)## is slightly

Textbook authors should know better than to use more decimals than their precision allows.

- #8

TSny

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Yes.Textbook authors should know better than to use more decimals than their precision allows.

- #9

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It is close enough to one of the answers to suspect that~~you (or~~the authors~~)~~have a rounding error somewhere.

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