A body is projected at 14.28 meters per second up the line of greatest slope of a plane inclined at an angle whose tangent is the square root of two over four. If the coefficient of friction between the plane and the body is three times the square root of two over five, what is the maximum distance that the body can travel up the plane? Take 𝑔 to equal 9.8 meters per second squared.
All right, so let’s say that this is our plane. And if we consider this angle of the plane, we’re told that its tangent is the square root of two over four. This is the ratio of the side opposite the angle to the side adjacent to it. So the base of our plane is four units long, while the height is the square root of two units tall. We’re further told that a body is projected at 14.28 meters per second up the incline of this plane. The coefficient of friction between the plane and the body is given. We want to figure out how far up the incline the body travels before it comes to rest. We’ll call this distance 𝑑.
Knowing that the acceleration due to gravity is 9.8 meters per second squared, let’s now clear some space on screen, and we’ll start on our solution by recognizing that as our body moves up this plane, both gravity as well as the force of friction will slow it down. Newton’s second law of motion tells us that the sum of forces acting on a body equal its mass times its acceleration. And as we look at the forces acting on our body, we can find similar information. We note first that there’s a gravitational force 𝑚 times 𝑔 acting on this body, and there’s also a normal or reaction force acting on it perpendicular to the surface of the plane.
And finally, as our body moves up the incline, a resistive force to the friction pushes the other way. We’ll label this force 𝐹. And we now have all three of the forces that act on this body as it moves up the incline. We want to solve for the body’s acceleration along this line of steepest slope. Let’s define the direction down the slope to be the positive 𝑥-direction. If we then consider all the forces that act along this dimension. The sum of those forces by Newton’s second law is equal to the mass of our body times its acceleration in the 𝑥-direction. Looking at our sketch of the forces acting on our body, we see that the frictional force 𝐹 acts in what we’ve called the positive 𝑥-direction and so does a component of the weight force that we’re highlighting here in orange.
To determine this component, we’ll want to know the measure of this angle here in our right triangle. If we return to our original sketch of the bottom of our screen, it turns out that this angle here is identical to the angle we’re indicating on our force diagram. To indicate this, we can give them both the same name. We’ll call this angle 𝜃. The component of the weight force that we’re interested in then is equal to 𝑚 times 𝑔 times the sin of 𝜃. Since this force acts in the same direction as our frictional force, we can add it to our sum of forces in the 𝑥-direction acting on our body.
But then we wonder what is the sin of this angle 𝜃? Returning once more to our original sketch, note that we have a right triangle where the length of the two shorter sides is square root two and four. By the Pythagorean theorem, the length of the hypotenuse ℎ of this triangle equals the square root of four squared plus the square root of two squared. This is the square root of 16 plus two or the square root of 18. But then we recall that 18 is equal to nine times two. The square root of nine is three. So a simplified expression for our hypotenuse length is three times the square root of two.
Now that we know the relative length of our triangles hypotenuse, we can solve for the sin of 𝜃. It’s equal to the side length opposite that angle divided by the hypotenuse length. We see here that the square root of two cancels from numerator and denominator so that the sin of 𝜃 is simply one-third. Therefore, 𝑚 times 𝑔 times the sin of 𝜃 can be written as 𝑚 times 𝑔 times a third. Along with this, it’s possible to write the frictional force 𝐹 in a different way. In general, for a moving body, the frictional force it experiences is equal to the coefficient of friction multiplied by the reaction force acting on the body.
In our scenario, our reaction force 𝑅 is equal in magnitude to this 𝑦-direction component of the weight force. This component is given by 𝑚 times 𝑔 times that cos of the angle 𝜃. So if this represents the reaction force magnitude 𝑅, we just need to multiply this by the coefficient of friction 𝜇 to arrive at an alternate expression for the frictional force 𝐹. And we’re able to do a bit more work on this expression because we can solve for the cos of 𝜃. It’s equal to the side length adjacent to the angle 𝜃, four, divided by the length of our hypotenuse, three times the square root of two.
So, this is our simplified expression for the frictional force 𝐹 that our body experiences. And if we add this to the weight force it experiences in the 𝑥-direction, then that sum is equal to the mass of our body times its acceleration along this dimension. Notice that the mass of our body 𝑚 appears in every term in this expression. This means if we divide both sides of the equation by mass, then it will cancel out from our entire equation. We can then factor the acceleration due to gravity from both terms on the left, and we can then plug in the given value for 𝜇 so we can solve for 𝑎 sub 𝑥 in terms of the acceleration due to gravity 𝑔. 𝑎 sub 𝑥 equals 𝑔 times four over three root two times three root two over five plus one-third.
We see that three times the square root of two cancel in this term for numerator and denominator. And then four divided by five plus one divided by three is equal to 17 divided by 15. So, this is the rate at which our body decelerates while it slides up the incline. This constant deceleration is vital information to help us solve for the distance 𝑑 it takes for our body to come to a rest. Because this body does change speed at a constant rate, its motion can be described by what are called the equations of motion. There are four of these equations in total, but we’ll use this particular one. It tells us that the final velocity of a body squared is equal to its initial or original velocity squared plus two times its acceleration multiplied by its displacement.
If we rearrange this equation algebraically to solve for the displacement 𝑑, we see it’s equal to 𝑣 sub 𝑓 squared minus 𝑣 sub zero squared all over two 𝑎. Now, in our case, we know the final velocity of our body is zero because we’re finding out how far it takes to come to rest. So, 𝑣 sub 𝑓 squared is simply zero. And we know 𝑣 sub zero. That’s given to us as 14.28 meters per second. Therefore, the numerator of our fraction looks like this. And we can now plug in for the acceleration 𝑎 that we solved for earlier. It’s equal to 𝑔 times 17 over 15. And then we can substitute in for 𝑔, which we know is 9.8 meters per second squared.
We’re just about ready to calculate 𝑑, but notice that if we do, we’ll get a negative result. This is actually consistent with the sign convention we’ve adopted since we’ve said motion down the incline is positive. Technically, we’re calculating a displacement which is different from a distance. Our problem statement, though, does ask for a distance rather than a displacement. To make the switch to calculating a distance, we only need to take the absolute value of our answer. When we calculate this fraction that way, we find a result of 9.18. And this distance is measured in units of meters. Our body then travels 9.18 meters up the slope before it comes to a stop.