Three spiders are resting on the vertices of a triangular web. The sides of the

Question

Three spiders are resting on the vertices of a triangular web. The sides of the triangular web have a length of a = 0.60 m, as depicted in the figure. Two of the spiders (S_1 and S_3) have +4.7 mu C charge, while the other (S_2) has -4.7 mu C charge. (a) What are the magnitude and direction of the net force on the third spider (S_3)? magnitude N direction degree counterclockwise from the +x-axis (b) Suppose the third spider (S_3) moves to the origin. Would the net force on the third spider (S_3) be greater than, less than, or equal to the magnitude found in part (a)? greater than in part (a) less than in part (a) equal to that in part (a) (c) What are the magnitude and direction of the net force on the third spider (S_3) when it is resting at the origin? magnitude N direction degree counterclockwise from the +x-axis

Explanation / Answer

first picture that point 3 is in the origin with the triangle out into quads 2 and 3, + + repulses so the force is directed into the 4th quadrant, - attracts and directs its force into the 3rd quadrant, so u can see it forms the same angle from the origin thus the X components cancel !!! so it just goes down the y axis (270 degrees) so all you have to do is use coulombs law to find the force , it should be the same for both

ok but now u have to find the force down, for the negative attractive charge looking at the triangle its 30 degrees, split the 60 degrees of the origin side , sin 30 * force = the first y component, the other should be the same so x2 and u get ur answer for magnitude

part c is easy they re in a straight line, like repels like neg attracts, thus they are going in the same direction, no x component , so just add the forces obtained by coulombs law, but remember that r is half what it was originally

hope that helps...

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