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Problem 29 Medium Difficulty

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are $m_{\mathrm{A}}=363 \mathrm{kg}, m_{\mathrm{B}}=517 \mathrm{kg},$ and $m_{\mathrm{C}}=154 \mathrm{kg}$ . Find the magnitude and direction of the net gravitational force acting on (a) particle A, (b) particle B, and (c) particle C.

Answer

$6.193 \times 10^{-5} \mathrm{N}$

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Video Transcript

There are two gravitational forces acting on particle. A one off the gravitational forces comes from particle B and another one from particle. See, So I disabled have be a there there additional force acting on particle A as an effect off the mass off particle Be similarly I labeled f c A. The magnitude off the gravitational force acting on particle A because off the mask off particle See, now we proceed to company to the magnitudes off these forces the magnitude off the force provoked by particle be on particle eight is equals to g times the mass off particle be times the mass off particle A divided by the distance between particle be and particle a squared, then plugging in their values that were given in the problem. We get that F B eight is a close to 6.67 times stand toe minus 11 times 363 which is the mask off particle 80 times 517 which is the mask off particle be divided by the distance between particles A and me and we can see in the figure that this distance is a close to 0.5 meters. So we have 0.5 squared down here and this gives us a gravitational force off approximately five 0.7 times 10 to minus five new terms. Now for the magnitude of the gravitational force between particle see and particle A, we have the following F c A busy Costa G M C times M A divided by the distance between particles, see and particle a squared. Then by plugging in the values that were given in the problem we get the following FC eight is equals to 6.6 to 7 times stand to minus 11 times 154 kilograms times 363 kilograms divided by the distance between particles A and C and the distance between particle A and C is he goes to 0.5 plus 0.25 which is 0.725 So we have 0.75 squared down here and this gives this a gravitational force off approximately six 0.6 to 9 times Stan to minus six mutuals. Now let me organize the board Now I have to choose a reference to calculate the net force So you choose that everything that is pointing to the right it's positive and as a consequence, everything that is pointing to the left is negative. The reform the net reputational force that is acting on particle A is given by, if be a plus f c A. Because both of these forces are pointing to the right and this Zico's to 5.7 times stand to minus five plus 6.6 to 9 times 10 to minus six, and this gives us a net force off. Approximately 5.6 to 7 times stand to minus five new tones on its pointing to the right. For the next item, we have to complete the magnitudes off the reputational forces that the particle a exerts on particle be and the magnitude off the gravitational force that part Cosi exerts on particle Be to begin with note the following the magnitudes off The reputational forces are symmetric. I mean that the magnitude off the gravitational force that particle a exerts on particle be Izzy coast to the magnitude off the reputation off works that particle be exerts on particle A. So we do not need to Copley these again. We had already calculated the magnitude off the gravitational force that particle be exerts on particle A. So we can just use this value again because off the symmetry on the magnitude stuff the gravitational force. So I'm not calculating it again. The only thing that we have to count leave this time is the magnitude of the gravitational force exerted on particle be by particle seat because we haven't completed it yet. Then we have the following f C V is it goes to g times the mass off particle C times the mass off particle be divided by the distance between particles B and C squared by the figure, you can see that the distance between particles B and C is equals to 0.25 meters. So by plugging the other values that were given by the problem, we get F C B has bean equals to 6.67 times stand toe minus 11 times 154 times 517 divided by 0.25 squared. And this gives us a gravitational force off a magnitude that is approximately pate 0.497 times stand to minus five Newtons. Now let me organize the Born before finishing Diz item. Okay, so the Net Force that acts on particle be easy Coz too f c b that is pointing to the right minus f a B that is pointing to the left and this is equals to 8.497 times stand to minus five miners. Remember that F A b is. It goes to f B eight and f B A is a question 5.7 times stand to manage five. So we have 5.7 times stand to minus five here and these gives us a net force off approximately three 0.49 times 10 to minus five new times to the right. Now let me clear the board for the next item on the last item we have to calculate the magnitudes off the reputational forced exert that on particle see by both particles A and particle. Be now remember that the magnitude of the gravitational force it's metric than F A C is equals two f c eight and we had already calculated f c A on the first item. It's here and also F B C is He goes to F. C V and had calculated FCB on the second item. Therefore, we do not need to copulate anti gravitational force again and we can proceed already directly to the net force. So the net force that acts on particle see is it goes to as both forces are pointing to the left. Both forces will have a minor sign the reform. We have miners f A, C minus F B C and this is a close to minus F A C, which is 6.6 to 9 times 10 to minus six minus F B C. We just 8.497 time stand to minus five and this gives us a net force off approximately miners. Nine points 16 times stand to minus five new terms. Off course we have a minus sign here. Therefore, this net force points to the left