Video Transcript
A small wooden box of mass 11 kilograms was placed at the top of a rough inclined plane of length 2.25 meters and height 1.8 meters. Given that it slid down the plane in one second, find the magnitude of its acceleration 𝑎 and the magnitude of the friction 𝐹 between the box and the plane. Take the acceleration due to gravity 𝑔 to equal 9.8 meters per second squared.
All right, let’s say that this is our inclined plane and that here is our box, starting at the top of the plane and ready to slide down. We know both the length of the plane as well as its height. And we also know that it takes one second for this box to slide all the way down the face of the plane. Knowing this, we want to solve for the box’s acceleration as well as the frictional force acting on it.
We’re given that the mass of the box, we’ll call it 𝑚, is 11 kilograms and that the acceleration due to gravity 𝑔 is 9.8 meters per second squared. Let’s now clear a bit of working space on screen. And we’ll begin on our solution by recalling Newton’s second law of motion, which tells us that the net force acting on an object equals that object’s mass times its acceleration.
Our next step will be to use this zoomed-in view of our box to draw in the forces that act on it. We know the box experiences a weight force, its mass times the acceleration due to gravity; a frictional force acting directly up the incline, we’ll call that 𝐹; and lastly a normal or reaction force that’s perpendicular to the plane. These are the forces acting on our box. And let’s say that the directions in this case are down the incline, that’s what we’ll call the positive 𝑥, and perpendicularly away from it, that’s what we’ll call the positive 𝑦.
If we then focus on forces in what we’ve called the 𝑥-direction, we can see that it includes a component of the weight force highlighted here, as well as the frictional force 𝐹 in its entirety. To figure out the magnitude of the weight force that acts in the positive 𝑥-direction, let’s go back to our original sketch. Looking at a right triangle, if we decide to call this angle here 𝜃, then we can see that the sine of that angle what we equal the opposite side of the triangle, 1.8 meters, divided by the hypotenuse, 2.25 meters.
Well, it turns out that this angle 𝜃 matches up with an angle in the right triangle for the components of our weight force. Returning to our free-body diagram up here, this angle in a right triangle is equal to 𝜃, which tells us that the component of the weight force acting in the positive 𝑥-direction is 𝑚 times 𝑔 times the sin of 𝜃. But then, as we’ve seen, the sin of 𝜃 is equal to this ratio, 1.8 to 2.25. We can say then that the total force pushing our box down the incline is given by this term. But then we know that that force is resisted by the force of friction 𝐹. Together, these forces add up to the box’s mass times its acceleration in the 𝑥-direction.
We can recall from our problem statement that 𝑎 sub 𝑥 and 𝐹 are the variables we want to solve for. Since we don’t yet know either of them, this equation won’t be enough by itself. That’s because it’s just one equation with two unknown quantities. However, notice that everything on the left-hand side of this expression, as well as the mass of the box, is a constant value. That means the box has a uniform acceleration 𝑎 sub 𝑥 as it slides down the plane. And any object that experiences a uniform acceleration is subject to what’s called the equations of motion.
In general, there are four equations of motion, but here we’ll just look at one. This equation, like all the others, only applies when an object is experiencing uniform or constant acceleration. The equation of motion we’ve chosen tells us that an object’s displacement is equal to its initial velocity times the time over which it is displaced plus one-half its acceleration times that time squared.
Let’s recall at this point that our box starts out at rest at the top of this plane. In other words, its initial velocity 𝑣 sub i is equal to zero. And we’re told further that the time it takes to reach the bottom of the incline is exactly one second. This means that we can write a much simplified version of this equation of motion for our particular scenario.
Since in our case 𝑣 sub i is zero and 𝑡 is one, we can just write that the total displacement of our box is equal to one-half its acceleration 𝑎. And note that this is an acceleration in what we’ve called the 𝑥-direction. From here, if we multiply both sides of this equation by two, that factor of two cancels with the factor of one-half on the right. And we find that 𝑎 sub 𝑥 is equal to two times 𝑠. Let’s recall that 𝑠 is the total displacement of our box. That equals the length of the face of our plane. So, leaving out the units, 𝑎 sub 𝑥 is equal to two times 2.25 or, in other words, 4.5. And its acceleration has units of meters per second squared. That then is our answer for the acceleration of this block, 4.5 meters per second squared.
And now, let’s substitute this value in for 𝑎 sub 𝑥 so that we can solve for the frictional force 𝐹. Leaving out the units for now, that substitution gives us this expression. And our task now is to rearrange this so that the friction force 𝐹 is the subject. We can do this by adding that force to both sides of the equation and subtracting 𝑚 times 4.5 from both sides. We end up with this expression. And if we factor out the mass of our box 𝑚, we can say that the friction force 𝐹 is equal to the box’s mass 𝑚 multiplied by 1.8 times the acceleration due to gravity 𝑔 over 2.25 all minus 4.5.
To solve for 𝐹, let’s now recall that the mass 𝑚 is 11 kilograms and the acceleration due to gravity 𝑔 is 9.8 meters per second squared. Plugging these in without their units gives us this expression. And entering this on our calculator gives us exactly 36.74. The SI base unit of force is the newton. So, we’ll say that our falling box experiences an acceleration of 4.5 meters per second squared while a frictional force of 36.74 newtons acts on it.
