WEBVTT
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A body of mass 16 kilograms was placed on a smooth plane that was inclined at 45 degrees to the horizontal.
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A horizontal force of 48 kilogram-weight was acting on the body towards the plane.
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Given that the line of action of the force, the body, and the line of greatest slope all lie in the same vertical plane, determine the acceleration of the body.
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Consider the acceleration due to gravity to be 𝑔 equals 9.8 meters per square second.
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To answer this question, we’re simply going to begin by sketching a diagram.
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This diagram doesn’t need to be to scale, but it should be roughly in proportion so we can accurately model what’s going on.
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We have a smooth plane inclined at 45 degrees to the horizontal.
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Now, the fact that this plane is smooth tells us simply that it exerts no frictional force on the body placed on it.
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We have a body whose mass is 16 kilograms placed on this plane.
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This means it exerts a downward force on the plane.
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And that force is equal to mass times acceleration due to gravity.
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We’ll call acceleration due to gravity 𝑔, although we are told it’s 9.8 meters per square second.
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The downward force the body exerts on the plane is 16𝑔.
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Now, we also have a horizontal force of 48 kilogram-weight acting on the body towards the plane.
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And so we can model this as the body exerting a horizontal force on the plane itself.
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The problem is we’re given information about this force in kilogram-weight.
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And currently, we’ve modeled the weight of the body to be in newtons.
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But really a kilogram-weight is simply another way of measuring a force.
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And we can convert from kilogram-weight to newtons by recognizing that one kilogram-weight is equal to 9.8 newtons.
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And so our horizontal force of 48 kilogram-weight is roughly equal to 48 times 9.8 newtons.
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That’s 470.4.
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So our horizontal force is 470.4 newtons.
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We’re looking to calculate the acceleration of the body.
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And so we need to make an assumption about the direction in which we think the body is going to move.
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Let’s assume it’s going to move up the plane.
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And so the acceleration is positive in this direction.
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Really, it doesn’t matter though.
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If we assumed the acceleration to be acting in the opposite direction, the sign would tell us the direction in which it really acts.
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Now, there’s one further force that we can model on our diagram.
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That force is the normal reaction of the plane on the body.
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Let’s call that 𝑅.
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Now, we don’t really need to consider 𝑅 in this question.
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But it’s always worth drawing on the diagram so we have everything modeled.
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And then the formula we’re going to use to help us to solve this problem is 𝐅 equals 𝑚𝑎: force is equal to mass times acceleration.
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We have modeled the acceleration as acting parallel to the plane.
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So we’re going to consider the components of our forces that act in this direction.
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Let’s begin by looking at our horizontal force.
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We’re going to add a right-angled triangle as shown.
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We know the included angle in this triangle is 45 degrees.
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And that’s because corresponding angles are equal.
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We want to find the component of this force that is parallel to the plane.
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So let’s label that side of our triangle 𝑥 or 𝑥 newtons.
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𝑥 is the adjacent side in this triangle, and the hypotenuse is 470.4 newtons.
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So let’s use the cosine ratio to form an equation.
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We know that cos 𝜃 is adjacent over hypotenuse.
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So in this case, cos of 45 degrees equals 𝑥 over 470.4.
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We can work out 𝑥 by multiplying by 470.4.
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So 𝑥 is 470.4 times cos of 45, but cos of 45 is root two over two.
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And so we find 𝑥 to be equal to 235.2 root two.
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That is the component of this force that acts parallel to the plane.
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We’ll now repeat this process for the weight force.
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We add a right-angled triangle in.
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And the included angle once again is 45 degrees.
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I’ve called the component of this force that’s parallel to the plane 𝑦 or 𝑦 newtons.
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Once again, we have the adjacent and hypotenuse.
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So we’re going to use the cosine ratio.
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cos of 45 is 𝑦 divided by 16𝑔.
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And if we multiply by 16𝑔, we get 16𝑔 times cos of 45.
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We’re now going to use the fact that 𝑔 is 9.8.
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Using this along the side the fact that cos of 45 is root two over two and we find 𝑦 is 78.4 root two.
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We now have forces acting parallel to the plane.
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We need to find the sum of these forces.
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This one acts in the same direction as the acceleration force, so we’re going to assume it’s positive.
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This one however acts in the opposite direction.
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It’s acting parallel to and down the plane.
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So its force is negative, meaning the sum of our forces is 235.2 root two minus 78.4 root two.
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This of course is equal to mass times acceleration.
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Remember, we’re using this formula.
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The mass of the body is 16 kilograms, and the acceleration we said was equal to 𝑎.
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So we have an equation: 235.2 root two minus 78.4 root two equals 16𝑎.
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This left-hand side simplifies to 156.8 root two.
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So to solve our equation, our last step is to divide through by 16.
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156.8 divided by 16 is 49 over five.
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So we find our acceleration 𝑎 is equal to 49 root two over five or 49 root two over five meters per square second.
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Note, now, that had we modeled the acceleration to be acting in the opposite direction — in other words, down the plane — we would have received a result of negative 49 root two over five.
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The magnitude of the acceleration though would have still been 49 root two over five.
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This just tells us that it’s decelerating in this direction.