Question Video: Finding the Kinetic Friction of a Body Moving on a Horizontal Plane with Uniform Acceleration by an Inclined Force | Nagwa Question Video: Finding the Kinetic Friction of a Body Moving on a Horizontal Plane with Uniform Acceleration by an Inclined Force | Nagwa

Question Video: Finding the Kinetic Friction of a Body Moving on a Horizontal Plane with Uniform Acceleration by an Inclined Force Mathematics • Third Year of Secondary School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

Using the information in the figure, calculate the coefficient of kinetic friction rounding the result to the nearest two decimal places given that the mass of the body is 28 kg and that the acceleration due to gravity 𝑔 = 9.8 m/s².

06:06

Video Transcript

Using the information in the figure, calculate the coefficient of kinetic friction, rounding the result to the nearest two decimal places. Given that the mass of the body is 28 kilograms and that the acceleration due to gravity, 𝑔, equals 9.8 meters per second squared.

Taking a look at our figure, we see our body which is on some surface, moving along with an acceleration to the right of 2.2 meters per second squared. This acceleration is due to a force of 155 newtons that’s being applied at an angle of 45 degrees to the horizontal. Knowing this, along with the fact that the mass of our body — we’ll call it 𝑚 — is 28 kilograms, we want to solve for the coefficient of kinetic friction between the body and the surface.

We can begin solving for this coefficient by first sketching in all the forces that are acting on our body. This is called a free-body diagram. For one thing, we know that our body is subject to the force of gravity. We also know that there’s an applied force, we’ll call it 𝐹 sub 𝐴, which is given as 155 newtons, acting at a 45-degree angle on the body. Furthermore, there’s a normal or reaction force acting straight up from the surface. And lastly, there’s a frictional force; we’ll call it 𝐹 sub 𝑓. And this opposes the motion of our object. That’s always true of friction, which is how we knew that it acted to the left.

Now that we know all the forces acting on our object, we can recall Newton’s second law of motion. This tells us that the net force that acts on somebody is equal to that body’s mass times its acceleration. Now because motion in the vertical direction is independent of motion in the horizontal direction, we can apply Newton’s second law to either one independently. In other words, we could say that the net force in the horizontal plane acting on our body is equal to the object’s mass multiplied by its horizontal acceleration. And indeed, this is the dimension we’ll focus on because we want to solve for the coefficient of kinetic friction, which is bound up in the force of friction.

When we say that, we just mean that the force of friction acting on an object that’s in motion is equal to the coefficient of kinetic friction multiplied by the normal force acting on the object. So, here’s what we’ll do. Focusing first on this horizontal direction, let’s clear a bit of space on screen and apply Newton’s second law in this horizontal plane. The first thing we’ll do is establish sign conventions, positive and negative.

Note that the acceleration given to us in our problem statement is a positive value and that acceleration is to the right. We’ll say then that any force or motion in that direction is positive, and any force or motion in the other direction is negative. As we study the forces acting horizontally in our free-body diagram, we see that there’s a component of the applied force that acts horizontally. And then there’s the friction force which acts entirely in this direction.

Given a 45-degree right triangle where the hypotenuse is our applied force magnitude, 155 newtons, we can solve for this horizontal leg of that triangle by multiplying 𝐹 sub 𝐴 by the cos of 45 degrees. That then accounts for the positive horizontal force acting on our body. We then subtract from that the frictional force. That force recall is equal to the coefficient of kinetic friction, what we want to solve for, times the normal force. This accounts for all of the horizontal forces acting on our mass. And therefore, by Newton’s second law, this sum is equal to mass times acceleration in the horizontal direction.

We’re given 𝐹 sub 𝐴 as well as the mass and acceleration of our body, but we don’t know 𝐹 sub 𝑁 to let us solve for 𝜇 sub 𝑘. To solve for it, we’re actually going to need to apply Newton’s second law again, but this time in the vertical dimension. As we do this, we can set up the conventions that motion up is in the positive direction and motion down is in what we’ll call the negative direction. So, let’s consider the vertically acting forces on our body. First, there’s the normal force, 𝐹 sub 𝑁. Added to that is the vertical component of our applied force, 𝐹 sub 𝐴.

Returning to our triangle, that side length is equal to 𝐹 sub 𝐴 times the sin of 45 degrees. And then lastly in the vertical direction, we have the weight force, 𝑚 times 𝑔, acting downward. By our convention, this force is negative. And by Newton’s second law, all of this is equal to the mass of our body times its acceleration in the vertical direction, what we’ll call 𝑎 sub 𝑣. We then realize that our object isn’t accelerating vertically, so this value is zero. That means these forces add up to zero. So, if we subtract this term from both sides of the equation and add this term to both sides, then we find an expression for the normal force, which is in terms of the weight force minus the vertical component of the applied force.

We can then substitute this whole right-hand side in for 𝐹 sub 𝑁 here. And now, we have an expression we can use to solve for 𝜇 sub 𝑘, the coefficient of kinetic friction. This is because we know the numerical values of 𝐹 sub 𝐴, 𝑚, 𝑔, and 𝑎, as well as the sin and cos of 45 degrees. Before we plug in those values though, let’s rearrange this expression so that 𝜇 sub 𝑘 is the subject. First, we’ll subtract 𝐹 sub 𝐴, cos of 45 degrees, from both sides. Then, we’ll divide both sides of the equation by this expression in parentheses. And lastly, we’ll multiply both sides by negative one. We end up with this expression. And now, we’re ready to substitute in our values.

𝐹 sub 𝐴 is 155 newtons. Both the cos and sin of 45 degrees is the square root of two over two. Our mass, 𝑚, is 28 kilograms. The acceleration, 𝑎, is 2.2 meters per second squared. And lastly, the acceleration due to gravity, 𝑔, is 9.8 meters per second squared. When we enter this expression on our calculator, to two decimal places, we find the result of 0.29. This is the coefficient of kinetic friction between our body and the surface.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy