Two uniform spheres, each of mass 0.260 kg, are fixed at points A and B (Fig. E13.5). Find the magnitude and direction of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.

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Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. To study Newton's law of gravity. A student fixes two metallic spheres, each of mass 5.0 kg separated by 24.0 centimeters. The student positions a third sphere of mass 0.50 kg at 5.0 centimeters from the midpoint of the line connecting the two spheres. The student releases the third sphere from rest. What is the acceleration magnitude and direction of the third sphere? If it experiences only gravitational force from the other two spheres treat all masses like point masses? So that's our end goal. Our end goal is trying to figure out what the acceleration magnitude and direction of the third sphere is if it only experiences the gravitational force and it comes from the other two spheres. So that's our final answer we're trying to solve is the acceleration magnitude and direction of the third sphere. Awesome. So below our problem, we're given a diagram to help us better visualize this problem, we have spheres one and two each labeled with a one and two respectfully. And they're all like represented by gray circles. And then we have our third sphere which is of a mass 0.50 kg. And the distance from our third sphere from the first sphere is 5.0 centimeters as shown with our two little folded black lines with arrows, knowing the distance and the distance between sphere one and sphere two, which as you can see, sphere one and spear two are significantly bigger circles than our third sphere. So the distance between sphere one and two is 24.0 centimeters. Awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. Let's also note that all the answers for the acceleration are in units of meters per second squared. And instead of saying direction each time, I'm just gonna say what direction it is, whether it's left or right. OK. So A is 3.95 multiplied by 10 to the power of negative eight. And left B is 3.95 multiplied by 10 to the power of negative eight and right C is 1.97 multiplied by 10 to the power of negative nine and right D is 1.97 multiplied by 10 to the power of negative nine and left E is 1.52 multiplied by 10. To the power of negative eight and left. And finally, for F it's 1.52 multiplied by 10 to the power of negative eight and right. Awesome. So keeping our diagram up up, so we can see it as we saw. First off, let us note that we are required to determine the acceleration magnitude and direction. So therefore, let us first draw a diagram to help us better visualize this problem. And let's make sure to label the forces the radius are in the angles. OK. So as you can see from our diagram, we can recall and use Newton's law of gravity and Newton's second law. So bef that's what we need to start thinking about, but really quick here. So looking at our third sphere, so to the far right of our little diagram here, so looking at our 0.50 kg sphere right through the center, going through the horizontal through the center of our third sphere, we have the angle from the horizontal that's directed towards our second sphere. And note that half of the distance because we know all the way across is 24 centimeters from sphere two to sphere one. So if the third sphere is at the midway point now just be half a 24 which would give us 12. And let us note that the vertical direction is the y direction and the horizontal direction is the X direction as according to our little coordinate plane here. Awesome. And our angle that's from the horizontal, from our third sphere is theta. OK. So with that in mind, we can now recall and use Newton's law of gravity and Newton's second law, which we could write as the sum which let's see, let's be very clear. So the sum of F is equal to FG one plus FG two is equal to mass multiplied by acceleration. OK. So let's note that this problem is symmetrical. Therefore, the magnitude of the forces from the two masses on the third mass will be equal. Also the Y components will cancel out. So considering this information, we can now write that the sum. So if we take this sum of FX, which is equal to negative F G one multiplied by the cosine of theta minus FG two. So FG two G two multiplied by cosine of theta is equal to mass multiplied by A X. OK. So the direction of the net force and the acceleration is negative and we also need to note that M one which we should be specific capital M one equals capital M two equals capital M. And let us use lowercase M for the third mass. And let us also note that cosine of theta is equal to X divided by R. And we can find the value of R by using Pgo Lois's theorem which states that R equals the square root of 5.0 centimeters square plus 12.0 centimeters squared, which when we perform that calculation, we will get 13.0 centimeters, which is also equal to 0.13 m. OK. So in case they're confused, like why are we using this theorem? So essentially we're imagining our third mass, our third sphere as like a triangle and we're using the triangle that's formed with. So basically our three spheres form like a triangle and we're treating the three spheres like as if they're in a triangular formation. And that's why we're able to use trig identities to help us solve for this problem or geometric properties to help solve for this problem. OK. So moving right along now that we know that we're dealing with a triangle, we need to solve for cosine of theta which is equal to X divided by R which is equal to 5.0 centimeters divided by 13.0 centimeters which is equal to 0.3846 when we round to four decimal places. So therefore, we can go ahead and write negative G multiplied by capital M multiplied by lower case M all divided by R squared multiplied by cosine of theta minus capital G multiplied by capital M multiplied by lower case M all divided by R squared multiplied by cosine of theta is equal to M multiplied by AX. So now we need to rearrange this equation to isolate and solve for AX because that's our final answer that we're trying to solve for, we're trying to solve for the acceleration. So AX is equal to the absolute value of negative two multiplied by G multiplied by capital M all divided by R squared multiplied by cosine of theta. Awesome. So now at this stage, we need to plug in all of our known variables and solve for the numerical answer of A X. So let's do that. So A X equals and to keep things simple, we all know that anything negative inside of absolute values inside the absolute value signs will become positive. So let's just make everything positive. So we'll just take two multiplied by 6.67 means that's our gravitational constant multiplied by 10 to the power of negative 11. And its units are meters cubed per kilogram multiplied by seconds squared, multiplied by 5.0 kilograms, all divided by 0.13 meters. Since we need to make sure we're using the proper units to make sure we everything cancels out. So 0.13 m squared and don't forget that everything is multiplied by 0.3 846 because that's our value for cosine of theta. So when we plug that into our calculator, we will find that A X is equal to 1.517 multiplied by 10 to the power of eight. And its units are meters per second squared which this is approximately equal to 1.52 multiplied by 10 to the power of negative eight. OK. We need to be specific. So when you plug in your calculator, you'll initially see 1.517 multiplied by 10 to the power of negative 8 m per second. But as we can see all of our multiple choice answers for the acceleration are rounded to three significant figures. So therefore, when we rounded three significant figures, we will get 1.52 multiplied by 10 to the power of negative 8 m per second squared. And that is our final answer for the acceleration hooray. We did it. So now to determine the direction, since the direction is negative, it will move to the left. So our direction is left, all right, we did it. So let's go look at our multiple choice answers to see which answer corresponds with the answers that we found together. And it appears that the letter E is our final answer. A equals 1.52 multiplied by 10 to the power of negative 8 m per second squared. And the direction is to the left. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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Key Concepts

Gravitational Force

Net Force and Acceleration

Vector Addition