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Math Help - Forces Test Redo

  1. #1
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    Forces Test Redo

    Hi. Our professor is letting us redo the questions we missed on our test for half-credit because the averages in his classes were 50, 60, and 70. I am looking for a bit of help on this question please.

    a.) Find the acceleration of Fnet = -5.0 N in the x direction - 3.0 N in the y direction, in magnitude angle form.

    b.) Find the acceleration of Fnet = 4.3 N in the x direction + 0.5 N in the y direction, in magnitude angle form.


    I cannot figure out how to figure that. Any help would be appreciated. Thanks!!
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  2. #2
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    Did those problems come with any masses since \Sigma F = mA. Once you've found the magnitude of net force by pythag, you can divide by the mass to find the acceleration and then use inverse tangent to find the angle. If there isn't a mass given then I'm not sure what the problem is asking. Hope that helps some.
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  3. #3
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    Yes, sorry. Overlooked that part of the question. Both masses are 5 kg. I guess what I am not sure is how to figure the acceleration in magnitude angle form since the net forces are in component form. FYI... the angle for a.) is 90 degrees and for b.) is 120 degrees.
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  4. #4
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    each force contributes to the motion of the object so i figure it looks some thing like the picture below

    \setlength{\unitlength}{2.5cm}<br />
\begin{picture}(100,100)(0,0)<br />
\put(1,-.05){x}<br />
\put( 0,1){y}<br />
\put( -.45,.1){-5 N}<br />
\put( .1,-.3){-3 N}<br />
\put(0,0){\circle{.1}}<br />
\put(0,0){\vector(-1,0){.5}}<br />
\put(0,0){\vector(0,-1){.3}}<br />
\put(0,0){\vector(-4,-3){.5}}<br />
\qbezier(0,0)(0,0)(0,-.9)<br />
\qbezier(0,0)(0,0)(-.90,0.00)<br />
\qbezier(0,0)(0,0)(.90,0.00)<br />
\qbezier(0,0)(0,0)( 0,.9)<br />
\end{picture}

    So the acceleration would be in the direction of the diagonal arrow, and the magnitude of Fnet by the pythagorean theorem is equal to \sqrt{x^2+y^2} and then A=\frac{\sqrt{x^2+y^2}}{m}. Also the angle must satistfy \sqrt{x^2+y^2}*\cos(angle)=x and \sqrt{x^2+y^2}*\sin(angle)=y, so angle=\arctan\frac{y}{x}. Does that help?
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