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If a body of mass three kilograms was placed on a smooth plane inclined at 17 degrees to the horizontal and was left to move freely, determine its acceleration to the nearest two decimal places.
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With questions like this, it can be really sensible to begin by sketching a diagram.
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Our plane is inclined at 17 degrees to the horizontal.
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Remember, this doesn’t need to be to scale.
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It’s just to give us an idea of what’s going on.
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A body of mass three kilograms is placed on this plane.
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And what happens is, this body exerts a downward force on the plane.
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The force is mass times acceleration.
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We said that the mass was three kilograms and the acceleration due to gravity is 𝑔.
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We will let that be equal to 9.8 meters per square second.
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But for now, we’ll leave it as 𝑔.
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Now, the fact that the body is placed on a smooth plane implies to us that there is no frictional force acting on the body by the plane.
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And so the only other force that acts on this body is the normal reaction force.
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This is the force of the plane on the body that acts at a perpendicular direction to the plane itself.
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It makes sense that the body when it’s released will want to roll down the plane.
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And so we mark acceleration in this direction as shown.
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So what next?
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Normally, we would begin by resolving forces perpendicular to the plane.
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But actually, we’re not interested in the normal reaction force.
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So instead, we’re going to resolve forces parallel to the plane.
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Now, the problem we have is that the weight of the body acts vertically downwards.
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That’s neither parallel nor perpendicular to the plane.
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And so we add a right-angled triangle.
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We’re now able to calculate the component of the weight that acts perpendicular to the plane.
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We’re going to call that 𝑥 or 𝑥 newtons.
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And then we also spot that we know the included angle in this triangle.
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It’s 17 degrees.
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And so since we’re trying to find the measure of the opposite side in this triangle, and we know the length of the hypotenuse, we can use right-angle trigonometry.
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We link the opposite side in the hypotenuse by using the sine ratio.
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So sin 𝜃 is opposite over hypotenuse.
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Substituting what we know about our triangle into this formula gives us sin of 17 degrees equals 𝑥 over three 𝑔.
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Remember, we’re trying to find the value of 𝑥, the component of the weight that acts parallel to the plane.
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So we’re going to multiply both sides of our equation by three 𝑔.
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And we find 𝑥 is equal to three 𝑔 sin 17.
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Now, once again, we could evaluate both 𝑔 as 9.8 and sin of 17 degrees.
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But we’re not going to just yet.
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Instead, we move on to the following formula: 𝐹 equals 𝑚𝑎.
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That is, force is equal to mass times acceleration.
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We know that the net sum of the forces that act parallel to the plane is three 𝑔 sin 17 degrees.
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Remember, there is no frictional force acting in the opposite direction since the plane is smooth.
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The mass of the body is three.
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And we’re trying to find the acceleration.
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Let’s call that 𝑎.
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So we get three 𝑔 sin 17 equals three 𝑎.
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We could divide through by three.
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And we’re now ready to evaluate 𝑎.
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It’s 𝑔 times sin 17.
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And we’ll replace 𝑔 with 9.8.
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Typing this into our calculator gives us 2.865 and so on.
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We’re going to round our answer to two decimal places.
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2.865 becomes 2.87.
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And we can therefore say that the acceleration of the body is 2.87 meters per square second.