Video Transcript
A body of mass 222 grams rests on a rough plane inclined to the horizontal at an angle whose tangent is four-thirds. The body is connected by a light inextensible string passing over a smooth pulley fixed at the top of the plane to a body of mass 310 grams hanging freely vertically below the pulley. Given that the coefficient of friction between the body and the plane is one-sixth, find the acceleration of the system. Take the acceleration due to gravity 𝑔 equals 9.8 meters per square second.
There is an awful lot of information here. So we’re going to begin by drawing a free-body diagram, that is, a diagram that just shows all of the key elements and the forces acting on each body. We’re told that a body of mass 222 grams rests on a rough plane inclined to the horizontal. This is inclined to the horizontal at an angle whose tangent is four-thirds. So let’s define that angle to be equal to 𝛼. And we can therefore say tan of 𝛼 is equal to four-thirds.
Let’s now add that first body to the plane. We’re told that the body has a mass of 222 grams, but of course the SI units for mass are kilograms. 222 grams is equivalent to 0.222 kilograms. This allows us to add the downward force of the weight of the body on that plane. The weight is given by the mass of the body in kilograms times 𝑔, the acceleration due to gravity. So the body therefore exerts a downward force on the plane of 0.222𝑔.
Then, we use Newton’s third law of motion. This tells us that since the body exerts a downwards force on the plane, the plane must exert a force in the opposite direction on the body. Now, in fact, we model this force as being perpendicular to the plane. Let’s call it 𝑅, and we’ll calculate the value of 𝑅 in a little while. Then, we’re told that the body is connected by a light inextensible string that passes over a pulley, which is fixed at the top of the plane, to a body of mass 310 grams that hangs freely vertically below the pulley as shown. Once again, we’ll convert this mass into kilograms, giving us 0.31. And so the downward force of the weight of this body must be 0.31 times 𝑔, where 𝑔 is once again the acceleration due to gravity.
Now, we’re not quite finished with the forces acting on each of these bodies. There is of course the tension force. On the first body, this acts in the direction parallel to and up the plane. And on the second body, it acts in the exact opposite direction of the downwards force of the weight. And in fact, we’re told that the coefficient of friction between the body and the plane is one-sixth. This means there is a frictional force that is trying to act against the motion of this object.
We might make the assumption that since this second particle has a greater mass, the first particle will attempt to move up the plane when the system is released from rest. And hence, the frictional force will act in the opposite direction. It will act in the direction down the plane. This frictional force 𝐹 sub 𝑟 of course can be calculated by finding 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the normal reaction force, as specified on our diagram. We are of course told that 𝜇 is equal to one-sixth here. And so we have all the information we need to start calculating the acceleration of the system. We define that to be equal to 𝑎. Let’s clear some space.
We’re going to begin by resolving forces on the first particle, the one on the plane. We want to do this because we need to calculate the value of the frictional force. And to do so, we need to calculate the value of 𝑅, the normal reaction force. So let’s begin by resolving forces in a direction perpendicular to the plane. Now, the sum of these forces is zero. In a direction perpendicular to the plane, the particle is itself in equilibrium. It is not leaving the plane, nor is it moving into the plane. So we have the normal reaction force that acts in this direction. But we also need to consider the force of the weight of this object. This acts in a direction neither parallel nor perpendicular to the plane. So we’re going to resolve this force into its two component parts.
To achieve this, we begin by drawing a right triangle as shown. The included angle in this right triangle is 𝛼. We’re trying to find the side in this triangle that lies adjacent to angle 𝛼. Since the hypotenuse is 0.222𝑔, this value is 0.22𝑔 times cos of 𝛼. Now, we don’t need to use the inverse trigonometric functions to find the value of 𝛼 here. If we recognize that four and three are two values in one of the recognized Pythagorean triples, we can draw a right triangle and find the value of cos of 𝛼 and sin of 𝛼, noting of course that this triangle is not to scale.
Since tan of 𝛼 is opposite over adjacent, our triangle might look a bit like this. The third side in this triangle is five units. Then, since the cosine ratio links the adjacent and the hypotenuse, we can say that cos of 𝛼 is equal to three-fifths. And since we’re here, sin of 𝛼 is opposite over hypotenuse; it’s four-fifths. So this component of the weight force is going to be 0.222𝑔 times three-fifths.
Let’s form an equation before we simplify this any further. We want to find the sum of the forces in this direction. So, if we take the direction away from the plane as being positive, the sum of the forces is 𝑅 minus 0.222𝑔 times three-fifths. And we said the particle is in equilibrium in this direction, so this sum is equal to zero. To solve for 𝑅, we add 0.222𝑔 times three-fifths to both sides of the equation. And then substituting 𝑔 equals 9.8 into this, we get 1.30536. So we have the normal reaction force of the plane on that first particle. Let’s add that to our diagram, and we’re now ready to resolve forces parallel to the plane.
Let’s begin by finding the component of the weight force of that first particle that acts in this direction. This time, that’s the side of the triangle opposite angle 𝛼. So it’s 0.22𝑔 times sin 𝛼, which we now know is 0.22𝑔 times four-fifths. This time, the particle is not in equilibrium; we know it’s moving. So we’re going to use 𝐹 equals 𝑚𝑎, and we’ll begin by finding the sum of the forces in this direction. Taking the direction that we have assumed this is going to move as positive, we have 𝑇 in the positive direction and then friction and 0.222𝑔 times four-fifths in the negative direction. So the sum of these forces is as shown. This is equal to the mass of this particle times acceleration, so 0.222𝑎.
Now, using the fact that friction is equal to 𝜇𝑅, and we have 𝜇 is equal to one-sixth and 𝑅 is equal to 1.30536. And then calculating the exact value of 0.222𝑔 times four-fifths, our equation becomes 𝑇 minus one-sixth times 1.30536 minus 1.74048 equals 0.222𝑎. And then that simplifies to 𝑇 minus 1.95804 equals 0.222𝑎.
We notice now that we have an equation involving two unknowns. So, if we can create another equation in 𝑇 and 𝑎, then we could solve those simultaneously. To achieve this, we resolve forces with our second particle. This is a little more straightforward. We’re simply working in the direction of gravity. If we take the positive direction to be once again the direction the particle is moving, the sum of the forces now are 0.31𝑔 minus 𝑇. Once again, this is equal to mass times acceleration, so 0.31𝑎.
We now have a pair of simultaneous equations. Remember, we’re trying to find the acceleration of the system, so it would be helpful to eliminate 𝑇 from our equations. Let’s rearrange our first equation to make 𝑇 the subject and substitute it into our second. That gives us 0.31𝑔 minus 1.95804 plus 0.222𝑎 equals 0.31𝑎. We subtract 1.95804 from 0.31𝑔, where 𝑔 is 9.8, and we get 1.07996. Distributing that negative one over the entire parentheses, and the left-hand side becomes 1.07996 minus 0.222𝑎.
Our next step will be to add 0.222𝑎 to both sides, giving us 1.07996 equals 0.532𝑎. And finally, we need to divide through by 0.532. That gives us 2.03. Now, that’s in meters per square second. And so we multiply through by 100 to get our acceleration in centimeters per square second. And so we’ve demonstrated the acceleration of the system is 𝑎 equals 203 centimeters per square second.
