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In this video, we’ll learn how to solve problems using Newton’s third law of motion.
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By this stage, you should’ve encountered and applied Newton’s first and second laws of motion.
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Newton’s first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
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And Newton’s second law says that the force acting on an object is equal to the mass of that object multiplied by its acceleration, 𝐹 equals 𝑚𝑎.
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So what does Newton’s third law tell us?
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Newton’s third law is most commonly stated as for every action, there’s an equal and opposite reaction.
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But we’re not talking about the type of reaction you have when, say, your sibling steals your favorite item of clothing.
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It actually says that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction on object A.
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As with Newton’s other laws, this is quite intuitive.
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For example, during a rocket launch, burning fuel releases gas at high velocity.
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The rocket exerts a downward force on the gas in the combustion chamber.
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And so the gas exerts an equal and opposite reaction force upward on the rocket.
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The special name for this is thrust.
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Another very important example applies to you at this very second.
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Whether you’re sitting on a chair or standing up, you are exerting a downward force on the chair or the floor.
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The chair or the floor is therefore exerting an upwards reaction force on you.
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This is the very thing that’s stopping you from falling straight through the chair or the ground, like in some sci-fi film.
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And so we have Newton’s third law of motion and some context behind it.
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Note that in classical mechanics, this law very rarely stands alone, and for this reason, you must make sure that you’re confident in applying the first two.
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Let’s begin then by applying the law in a very simple example.
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An elevator is moving vertically upward at a constant speed.
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A man of mass 150 kilograms is standing inside.
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Determine the reaction force of the floor on the man.
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Take 𝑔 to be equal to 9.8 meters per square second.
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With questions like these, it can be really helpful to begin by drawing a very simple diagram.
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Here’s the lift moving upward at a constant speed.
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Now, if the speed is constant, acceleration must be equal to zero meters per square second.
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A man of mass 150 kilograms stands inside the lift.
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And so this means that the man must exert a downward force on the floor of the lift.
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Now, in fact, to find the value of that downward force, we can use Newton’s second law, force is equal to mass times acceleration.
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We call the downward force due to the mass “weight”, 𝑤.
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And we say it’s equal to the mass of the object multiplied by acceleration due to gravity, so 𝑚𝑔.
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This must mean that since the man has a mass of 150 kilograms, the downwards force of weight is 150𝑔 on the floor of the lift.
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Then we recall Newton’s third law of motion.
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And this says that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction on object A.
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And so since the man exerts a downwards force on the floor of the elevator, the floor of the elevator must exert an upward force on the man itself.
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And that’s all the forces we’re interested in for now.
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We aren’t interested in, for example, tension forces in the cable for the lift.
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And so we’re going to go back to Newton’s second law of motion, force is equal to mass times acceleration.
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We’re going to define upwards in this case as being positive, and we’re going to find the net sum of the forces.
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In the upwards direction, we have 𝑅, and then downwards, we have 150𝑔.
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This is acting in the negative direction.
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And so the net force on our diagram is 𝑅 minus 150𝑔.
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This is equal to mass times acceleration, 150 times zero.
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150 times zero is zero.
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So our equation becomes 𝑅 minus 150𝑔 equals zero.
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We can solve for 𝑅 by adding 150𝑔 to both sides, and we find 𝑅 is 150𝑔.
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But of course we were told 𝑔 is 9.8, so we substitute 𝑔 equals 9.8 into this expression.
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And we get 𝑅 equals 150 times 9.8, and that’s 1470.
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We’ve worked with kilograms and meters per square second, so the units for our force are newtons.
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The reaction force of the floor on the man is 1470 newtons.
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Now it’s worth noting that we could have gone straight to this step.
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We didn’t actually need the first line of working.
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It’s essentially the reverse of the first law.
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We’re told that an object will remain at rest or in uniform motion unless acted upon by an external force.
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In other words, if the sum of the forces acting on the body is zero, then the object will remain at rest or in uniform motion.
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Now we know that our object remains in uniform motion, and so the sum of the forces did have to be equal to zero.
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We’ll see why, though, in our next example, it can be really helpful to use Newton’s second law most of the time.
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An elevator was accelerating vertically downward at 1.7 meters per square second.
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Given that the acceleration due to gravity is 𝑔 equals 9.8 meters per square second, find the reaction force of the floor to a passenger of mass 103 kilograms.
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Let’s begin by sketching a diagram.
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We have an elevator accelerating vertically downward at 1.7 meters per square second.
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We know that a passenger in the lift has a mass of 103 kilograms.
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This passenger must exert a downward force on the lift, and we call that downward force weight, where weight is mass times 𝑔 gravity.
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In this case then, since the passenger has a mass of 103 kilograms, his weight is 103𝑔.
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Newton’s third law then tells us that if an object exerts a force on a second object, that second object must exert a force of equal magnitude and opposite direction on the first.
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And so the floor of the lift exerts a reaction force on the passenger itself.
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We now have all of the forces that we’re interested in, and so we recall Newton’s second law of motion, force is equal to mass times acceleration.
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We’re going to find the net sum of the forces in our diagram, and we’re going to set that equal to the mass of the passenger multiplied by acceleration.
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And so we’re going to need to define a positive direction.
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Now, this is very much personal preference.
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I prefer to take the positive direction as the direction at which the object is moving.
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So in this case, I’m going to assume that downwards is the positive direction.
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As long as we’re consistent, though, it doesn’t matter which direction we choose.
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So forces acting in the positive direction, we have 103𝑔.
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In the negative direction, we have 𝑅, and so the net sum of the forces in this diagram is 103𝑔 minus 𝑅.
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That’s equal to the mass of the passenger multiplied by acceleration, so 103 times 1.7.
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Remember, we’re trying to find the reaction force of the floor to the passenger, so we need to make 𝑅 the subject.
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Let’s begin by adding 𝑅 to both sides.
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We could, of course, have multiplied by negative one instead, but we do need 𝑅 to be a positive.
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Then we’re going to subtract 103 times 1.7 from both sides, and so we get 103𝑔 minus 103 times 1.7 equals 𝑅.
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It can be sensible to factor where possible.
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This makes any mental calculation a little easier.
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And the left-hand side then becomes 103 times 𝑔 minus 1.7.
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But, of course, 𝑔 is equal to 9.8, and so since 9.8 minus 1.7 is 8.1, we find 𝑅 must be equal to 103 times 8.1.
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That’s 834.3.
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We have been working with kilograms and meters per square second, and so the units for our force are newtons.
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The reaction force of the floor to a passenger of mass 103 kilograms is 834.3 newtons.
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Now, comparing this reaction force to the weight force of the man might remind you of a time where you’ve been in a lift yourself.
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The weight force 103𝑔 is 103 times 9.8, and that’s 1009.4.
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1009.4 is greater than the reaction force, and this actually makes a lot of sense.
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If you’ve ever been in a lift which accelerates downwards, you almost feel a little bit lighter than normal.
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Similarly, if you’re in an elevator accelerating upwards, you might feel a little bit heavier than normal.
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And this is because the reaction force will be greater than the weight force.
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In our next example, we’re going to consider how to work with a spring balance.
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A body of mass 32 kilograms was suspended from a spring balance fixed to the ceiling of an elevator.
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Given that the elevator was accelerating upward at 405 centimeters per square second, find the apparent weight of the body.
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Take 𝑔 to be equal to 9.8 meters per square second.
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Let’s begin by sketching a diagram.
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We have a body of mass 32 kilograms suspended from a spring balance.
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That spring balance is fixed to the ceiling of an elevator.
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And so there are a couple of forces at play here.
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Firstly, there’s the downwards force of the body.
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That force is the weight, and it’s equal to mass times gravity, so 32 times 𝑔.
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Now, since the body itself is exerting a downward force on the spring balance, Newton’s third law tells us that the spring balance must also exert an upward force on the body itself.
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Let’s call that force tension, and actually that tension will give us the reading and the apparent weight of the body.
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We know that the elevator itself is accelerating upward at 405 centimeters per square second.
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But actually, since there are 100 centimeters in a meter, we’re going to divide this value by 100 to get 4.05 meters per square second.
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And that’s because we’re currently working in kilograms.
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We’ve got gravity in meters per second, and so our final answer is going to be in newtons.
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And so we need to be consistent with our units throughout.
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Now, actually, the spring balance itself will exert a force on the ceiling of the elevator.
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But we’re not interested in that system of forces.
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And so we move on, and we use Newton’s second law of motion, force is equal to mass times acceleration.
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We’re going to work out the net sum of the forces in our system, and then we’re going to set that equal to the mass of the body times the acceleration.
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And so we need to choose a positive direction.
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Let’s take upwards to be positive, since that’s the direction in which the body is moving.
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And since we have 𝑇 tension acting upward and 32𝑔 acting in the negative direction, the net sum on our system is 𝑇 minus 32𝑔.
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That’s equal to the mass of the body times acceleration.
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So that’s 32 times 4.05.
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We’re trying to find 𝑇 since that will tell us the apparent weight of the body.
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And so we’re going to add 32𝑔 to both sides.
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𝑇 is 32 times 4.05 plus 32𝑔, which is equal to 32 times 4.05 plus 32 times 9.8.
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That gives us a value of 443.2.
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The apparent weight of the body then is 443.2 newtons.
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In our final example, we’ll look at how we can find acceleration by using a system of simultaneous equations.
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A body is suspended from a spring balance fixed to the ceiling of an elevator.
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The reading of the balance was 50 kilogram-weight when the elevator was accelerating upward at 𝑎 meters per square second, and the reading was 10 kilogram-weight when the elevator was accelerating downward at five-thirds 𝑎 meters per square second.
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Given that the acceleration due to gravity is 𝑔 equals 9.8 meters per square second, determine the value of 𝑎.
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Let’s begin by sketching a diagram of each part of this scenario.
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Initially, the elevator is accelerating upward at 𝑎 meters per square second.
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The reading of the balance at this point was 50 kilogram-weight.
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Try not to worry too much about the units here.
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Kilogram-weight is just another way of measuring force, and so we simply write 50 as the reading on our balance, and that’s essentially the upward force.
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That upward force is a reaction to the downward force of the weight of the body.
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Now we don’t know the mass, so we’ll let that to be equal to 𝑚.
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And the downward force is mass times gravity; it’s 𝑚𝑔.
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We then have a reading of 10 kilogram-weight when the elevator is accelerating downward at five-thirds 𝑎 meters per square second.
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The downwards force of the weight 𝑚𝑔 remains unchanged.
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And so now we have the relevant forces.
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Let’s use Newton’s second law, force is equal to mass times acceleration, taking upwards to be positive in our first diagram and the net force is 50 minus 𝑚𝑔.
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That’s equal to mass times acceleration 𝑚𝑎.
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Then taking the downwards direction to be positive in our second diagram, the sum of the forces is 𝑚𝑔 minus 10.
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That’s equal to mass times the new acceleration, 𝑚 times five-thirds 𝑎.
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Note that we could have chosen upwards to be positive.
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It really doesn’t matter.
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We’ll get the same result.
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We rewrite the right-hand side as five-thirds 𝑚𝑎, and the question tells us that we’re trying to find the value of 𝑎.
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So we’re going to make 𝑚 the subject in each of our equations, and then we can set them equal to one another.
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In our first equation, we add 𝑚𝑔 to both sides, and then we factor.
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So we get 50 equals 𝑚 times 𝑎 plus 𝑔.
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Finally, we divide by 𝑎 plus 𝑔, and so 𝑚 is 50 over 𝑎 plus 𝑔.
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In our second equation, we subtract five-thirds 𝑚𝑎 and add 10.
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We factor once again, and we’re going to divide by the expression inside our parentheses.
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𝑚 is 10 over 𝑔 minus five-thirds 𝑎.
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Now that we have two equations for 𝑚, we’re going to set them equal to one another.
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50 over 𝑎 plus 𝑔 equals 10 over 𝑔 minus five-thirds 𝑎, and then we divide through by 10.
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We’re going to multiply by both denominators.
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That gives us five times 𝑔 minus five-thirds 𝑎 equals 𝑎 plus 𝑔.
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Let’s distribute the parentheses and then make 𝑎 the subject.
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Now, of course, 𝑔 itself is a number; it’s 9.8.
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Adding 25 over three 𝑎 to both sides and subtracting 𝑔, and we get four 𝑔 equals twenty-eight thirds 𝑎.
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But of course, four 𝑔 is four times 9.8, which is 39.2.
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Our final step is to divide both sides of this equation by twenty-eight thirds.
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39.2 divided by twenty-eight thirds is 4.2, and so the acceleration 𝑎 is 4.2 meters per square second.
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Let’s now recap the key points from this lesson.
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In this lesson, we learned that Newton’s third law of motion says that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction on object A.
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We also saw that this law is very rarely used alone.
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It will be used quite commonly alongside the first and second laws for motion.