In an electricity experiment, a 1.0 g plastic ball is suspended on a 60-cm-long string and given an electric charge. A charged rod brought near the ball exerts a horizontal electrical force F(electric) on it, causing the ball to swing out to a 20 degree angle and remain there. (a) What is the magnitude of F(electric)?

Video transcript

Hey, everyone in this problem, we have an experiment in which a 40 centimeter long thread suspends a bar magnet with a mass of 10 g. When another bar magnet is brought close to the first one with the identical poles facing each other. OK. Either north to north or south to south, they're gonna repel one another. We're told that if the magnet is held in place, OK. That second magnet, it causes the hanging magnet to swing out to a 15 degree angle and remain there. And you can see that in the diagram we shown and we're asked to determine the magnitude of the magnetic force that causes the repulsion between magnets and that makes it swim. And again, we're given a diagram when the magnet is just hanging and then when it moves to that 15 degree angle, once it repels from the other magnet, we're given four answer choices. All in Newton's option A 0.45, option B 0.3, option C 0.26 and option D 0.8. We're gonna scroll down here and we're gonna first start by drawing a free body diagram. OK? We're asked for the magnitude of a particular force. So when we think about forces, we want to think about our free body diagram. So we have our magnet, we know that we always have the force of gravity pointing straight downwards, right. That is F G, we have this force pointing to the right. And we're gonna call that F M and that is gonna be that magnetic force that we're looking for. OK. From that repulsion, the other forces we have acting is tension from this string or this cable. And that is pointing up into the left and we know that that makes a 15 degree angle with the vertical, we can break this into the horizontal and vertical components. OK? So we have the tension T, we have the vertical component T Y and we have the horizontal component T X A and again, the angle with the horizontal or sorry with the vertical is 15 degrees. All right, let's take up into the right as our positive directions. What we wanna find is this force F N and you know, if we look in the horizontal direction, we have this force F M and we also have the force T X. OK? And we don't know what T X is. So in order to look in the horizontal direction and calculate F M, we first need to figure out T X. Yeah, which means we're gonna look in the vertical direction in that Y direction to find T Y so that we can relate it to T X. OK. So starting with the forces in the y direction and we know that this is gonna be equal to zero. OK. Why is that? Well, we're told that this magnet moves out to a 15 degree angle and then remains there. OK. So that magnet is not moving, it's an equilibrium so that some of the forces are gonna be equal to zero in the positive Y direction. We have that tension T Y in the negative Y direction, we have the force of gravity F G. So we get T Y minus F G is equal to zero. OK. T Y. We can write in terms of the tension T and it's gonna be related to the cosine of the angle because we're talking about the adjacent side. So we have t multiplied by cosine of 15 degrees. We can move the force of gravity F G to the right by adding it. And recall that F G, the force of gravity is given by the mass multiplied by the gravitational acceleration G isolating for tea, we divide by cosine of 15 degrees. We're gonna substitute in our mass. And we're told that this bar mass has a mass of 10 g. OK. Now our answers are all in newtons, we're called that a Newton is a kilogram meter per second squared. So we need to convert this mass into a kilogram. So that we end up with that unit of Newton that we want. We're gonna multiply and in every one kg, we know there are 1000 g who multiply by one kg divided by 1000 g, the unit of gram divides out. OK. We're left with kilograms. What we're essentially doing here is dividing by 1000. This is multiplied by the acceleration due to gravity 9.8 m per second squared. And all of this is divided by cosine of 15 degrees. Now, we can simplify the numerator. We get that the tension T is gonna be equal to 0.98 newtons divided by cosine of 15. And we're gonna leave it just like that for now. So we don't have to do any unnecessary rounding, leave it like that. And now we're gonna move to the X direction. OK. The horizontal component so that we can find that magnetic force F M. Now the sum of the forces in the X direction again is gonna be equal to zero because this magnet is not moving. It's an equilibrium in the positive X direction. We have that magnetic force F F in the negative X direction. We have the tension T X. OK. So F M minus T X is equal to zero. This tells us that that magnetic force is equal to the tension force in the X direction. OK? We can add T X to the right hand side and we can break that up just like we did for the vertical component. This is gonna be related through the angle with sine because we're talking about the opposite side. Now. So we have the tension T multiplied by sine of 15 degrees. Now, we know what tea is. We just calculated tea so we can substitute tea from our vertical equation into our horizontal K T is gonna be 0.98 newtons divided by cosine of 15 degrees. And that is multiplied by sign of 50 degrees. We can simplify a little bit. OK? We know that sine of 15 degrees divided by cosine of 15 degrees will give us tangent of 15 degrees. So we can write this as 0.98 newtons multiplied by tangent of 15 degrees. And if we use our calculator to approximate this value, we get that this force is gonna be equal to 0.26, 26 noons. And that is that magnetic force that is causing the repulsion that we were looking for. If we compare this to our answer choices, we can round to two decimal places like the answer choices have here. We can see that our answer corresponds with answer choice B 0.3 newtons. Thanks everyone for watching. I hope this video helped see you in the next one.