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A piano of mass 300 kilograms is moved along a horizontal surface by a person pushing from one side and another person pulling from the other.
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The piano accelerates at 0.25 meters per second squared.
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The pushing force is 120 newtons, and a friction force of 60 newtons acts in the opposite direction to the piano’s velocity.
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What is the pulling force?
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Consider the direction in which the piano moves to be the positive direction.
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Okay, so in this question, we know that we’ve got a piano which has a mass of 300 kilograms and we know that it is moved along a horizontal surface by two people: one person is pushing from one side and the other person is pulling from the other side.
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So let’s draw a diagram showing the situation.
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Here’s our piano then.
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We know that the piano has a mass of 300 kilograms and we’ll call this mass 𝑚.
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Now, another piece of information we have is that the piano accelerates at 0.25 meters per second squared.
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So let’s say that the piano is travelling from left to right.
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And so its acceleration is 0.25 meters per second squared from left to right.
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As well as this, we’ve already accounted for the fact that the piano is moving along a horizontal surface because this is how we’ve drawn it.
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It’s moving from left to right and so it’s moving on a flat plane.
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Now, we’re told that the pushing force the person pushing is applying a force of 120 newtons.
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And because the piano is travelling from left to right, naturally, they’ll be pushing in this direction from left to right as well.
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Let’s call this force 𝐹 sub one, which is 120 newtons.
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As well as this, we’re told that a friction force of 60 newtons acts in the opposite direction to the piano’s velocity.
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So there’s going to be a friction force along the ground to the left.
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Let’s call this 𝐹 sub fric for friction and we’ve been told that this friction force is 60 newtons.
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Now, what we’ve been asked to do is to find out the pulling force.
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In other words, there’s a second person pulling from the right-hand side of the piano and so they’ll be pulling in this direction.
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Now, we don’t know what this force is.
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So let’s call this force 𝐹 sub two and put a question mark next to it.
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As well as this, it’s also worth labelling the acceleration of the piano as 𝑎.
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And so at this point, we’ve labelled all of the forces that we know as well as the mass of the piano, the acceleration of the piano, and the force that we’re trying to find, which is 𝐹 sub two.
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Now, the final sentence in the question is giving us a convention to work with.
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It’s telling us that the direction in which the piano moves should be the positive direction.
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So we say that if we’re going towards the right, we’re going in a positive direction and vice versa — negative direction to the left.
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So we’ve got to this point now, let’s start working out what 𝐹 sub two is.
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Now, in our diagram, we’ve got all the forces acting on the piano.
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What we can do with this information is to work out the net force on the piano.
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Now, this is the overall force and this accounts for all the individual forces acting on the piano, whilst also considering the directions in which they act.
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So the net or resultant force on the piano, which we’ll call 𝐹 sub net, is equal to firstly 𝐹 one which is 120 newtons plus 𝐹 two which is the pulling force that we’re trying to find out minus 𝐹 sub fric.
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And the reasoning for this is that both 𝐹 one and 𝐹 two act towards the right.
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So they act in the positive direction, whereas 𝐹 sub fric acts toward the left in the negative direction.
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So we stick a negative sign in front of 𝐹 sub fric.
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And so this expression is going to give us the net force on the piano.
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Now, why is this useful?
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Well, we’ve been given the acceleration and the mass of the piano.
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As well as this, we’ve just worked out an expression to give us the net force on the piano.
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And there’s a very handy relationship known as Newton’s second law of motion that we can use.
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Newton’s second law of motion tells us that the net force on an object 𝐹 sub net is equal to the mass of that object multiplied by its acceleration.
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What this means is that we can use Newton’s second law and substitute it in the right-hand side of this expression instead of 𝐹 sub net.
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And on the right-hand side, we can write down the mass multiplied by the acceleration coming straight down from here.
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Now, at this point, we’ve got a relationship between forces that we know — specifically 𝐹 one and 𝐹 fric — as well as the mass of the piano and the acceleration which we also know.
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And we’ve got one unknown variable; that is, 𝐹 two, which is what we’re trying to find out.
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So we can rearrange this equation for 𝐹 two which we can do by subtracting 𝐹 one and adding 𝐹 fric to both sides of the equation.
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Doing this means that 𝐹 one minus 𝐹 one cancels out and minus 𝐹 fric plus 𝐹 fric also cancel out.
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And that just leaves us with 𝐹 two on the left-hand side.
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And at this point, we can substitute in all the values we’ve been given in order to find out what 𝐹 two is.
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So we’ve got 𝐹 two is equal to the mass 300 kilogram multiplied by the acceleration 0.25 meters per second squared minus 𝐹 one which is 120 newtons plus 𝐹 fric which is 60 newtons.
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And we’ve used standard units for all of the quantities in this equation.
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The standard unit for mass is kilograms, which is what we have the mass in.
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For acceleration, we have metres per second squared also the standard unit.
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And we also have the two forces in newtons which is their standard unit.
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So our final answer of 𝐹 two is going to be in its standard unit as well, which is newtons.
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And once we evaluate the right-hand side of the equation, we find that the value of 𝐹 sub two is 15 newtons.
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And at this point, we have our final answer: the pulling force 𝐹 two is 15 newtons.