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