Lesson Video: Normal Reaction Force on a Sloping Surface Physics

Video Transcript

In this video, we will learn how to analyze weight, friction, and normal reaction forces contributing to the motion of objects on sloped surfaces. Before we move on to sloped surfaces, let’s first refresh our memory about using these forces on a horizontal surface.

Let’s begin by drawing all the forces acting on our block which is sitting on a horizontal surface. The weight of the block, which we will represent as 𝑊 in this video, originates from the center of mass. And the normal reaction force, which we will call 𝑅 for this video, which originates from the surface at a position that’s directly underneath the center of mass. We should remember that the normal reaction force will always make a 90-degree angle with the surface or be normal to the plane.

In this situation where an object is sitting on a horizontal surface, the weight of the object will be in the opposite direction to the normal reaction force. In fact, in the absence of any other vertical forces or vertical component of forces, the weight of the object will be equal to the negative of the normal reaction force, meaning they will have the same magnitudes but will be acting in opposite directions. If we were to alter the situation by trying to pull on our block, we would have to include our force of friction, assuming that the surface is rough. Our force of friction will pull in the opposite direction to our pull force. In this case, our force of friction will be to the right of our screen and will be represented by 𝐹 subscript f.

If our object remains at rest, the negative 𝑃 will equal 𝐹 f, which means that the pull force and the force of friction will have the same magnitude but will be heading in opposite directions, as can be seen in the diagram with the pull force to the left and the force of friction to the right. If we draw in black dotted lines to represent the horizontal and vertical axes, we would see that all four of the forces align with the coordinate system. In reference to the plane which happens to be along the horizontal axis, we can see that the pull force and the force of friction are parallel to the plane. Being perpendicular to the plane would be in the same direction as the vertical axis such as the normal reaction force and the weight of our object.

Let’s apply these concepts to an object that is on a sloped surface. Being a sloped surface means that the plane has been rotated with reference to the coordinate system. In this case, it was rotated by an angle 𝜃. If we assume that we put the box on an inclined plane with a smooth surface, the weight of the object will still be pulling straight down along the vertical axis, starting at the center of mass of the object. And the normal reaction force is perpendicular to the surface directly below the center of mass. This can be drawn in one of many ways. In this video, we have chosen a common convention to draw the normal reaction force through the center of mass. Another way is to start the force on the plane where the force of gravity intersects the plane and draw the normal reaction force perpendicularly.

Many times we use a simple point to represent our object to avoid confusion when drawing a free-body diagram. Unlike on a horizontal surface, on a sloped surface, we can see that our normal reaction force does not align with the vertical axis. The force would have to be split into its two components. We could call them 𝑅 subscript V for the vertical component and 𝑅 subscript H for the horizontal component. For this example, the vertical component would be pointing to the top of the screen, which is in the opposite direction as the weight of the object. And the horizontal component would be pointing to the left of the screen.

By keeping the coordinate system as vertical and a horizontal, it makes it difficult to analyze the motion of the object. Instead, we rotate our coordinate system so that an axis is perpendicular to the plane and an axis is parallel to the plane. Now the weight needs to be broken down to its components so that it is aligned appropriately with the coordinate system instead of their reaction normal force. The two components are the weight perpendicular, which is the component of the weight that’s perpendicular to the plane and opposes the normal reaction force, and the weight parallel or the component of the force that is parallel to the plane. We have used subscripts that have the symbol for perpendicular and parallel in order to tell these two components apart.

If our block has a mass 𝑚, let’s determine what the perpendicular and parallel components of the weight would be in terms of known constants and variables. The weight of the object and its components make a right triangle, with the angle at the top here being 𝜃, the same as the angle that our surface was sloped at. The parallel component of the weight is the opposite side of the triangle to the angle 𝜃, and the weight is the hypotenuse of the triangle. Remembering trigonometry functions SOHCAHTOA, we can use SOH as we have opposite and hypotenuse. The sin of the angle 𝜃 will equal the opposite side of the triangle divided by the hypotenuse.

Plugging in our variables, we have the sin of 𝜃 will equal the parallel component of the weight divided by the weight. To isolate the parallel component, we multiply both sides by 𝑊. The weight of the object times the sine of the angle is equal to the parallel component of the weight. We should remember that the weight of an object can be found by multiplying the object’s mass by the acceleration due to gravity. So we could replace 𝑊 in our equation with 𝑚𝑔, leaving us with the parallel component of the weight being equal to 𝑚𝑔 sin 𝜃.

Let’s use the same approach to determine the perpendicular component of the weight. The perpendicular component of the weight is the adjacent side of the triangle to 𝜃, and once again the weight is the hypotenuse. From our trigonometry functions, we can use the cosine. cos of 𝜃 is equal to the adjacent side of the triangle divided by the hypotenuse. Plugging in our variables, we have the cos of 𝜃 is equal to the perpendicular component of the weight divided by the weight. To isolate the perpendicular component, we multiply both sides by 𝑊. The weight of the object times the cos of the angle 𝜃 is equal to the perpendicular component of the weight. Once again, since weight is equal to 𝑚𝑔, we can replace 𝑊 with 𝑚𝑔 and get 𝑚𝑔 cos 𝜃 is equal to the perpendicular component of the weight.

Let’s analyze the perpendicular forces and parallel forces to determine the motion of our block. To do this, we’ll apply Newton’s second law 𝐹 net equals 𝑚𝑎 to the perpendicular axis and to the parallel axis separately. The two forces that we add together along the perpendicular axis to get the net force are the normal reaction force and the perpendicular component of the weight. We have chosen the common convention as into the plane of being negative, which is why 𝑅 is positive and 𝑚𝑔 cos 𝜃 is negative. Our block is not accelerating into or away from the surface of the plane, and therefore 𝑚𝑎 will be zero. So the magnitude of the normal reaction force is equal to 𝑚𝑔 cos 𝜃.

Let’s look at the extreme cases of a sloped surface, a horizontal surface and a vertical surface. We did the horizontal surface earlier, and we saw that 𝑅 is equal to 𝑊, assuming there are no other vertical forces. Looking back at our formula, that should make sense as the cos of zero degrees is one. When our plane makes an angle 𝜃 of zero degrees to the horizontal, then the reaction normal force is equal to the weight or 𝑚𝑔. When our block is sliding along the vertical surface, the angle 𝜃 that the plane makes with the horizontal would be 90 degrees, and the cos of 90 degrees is zero. So the normal reaction force in this case is equal to zero. Assuming a smooth surface, the only force acting on the object would be the weight, so the object would essentially be in free fall.

Next, we’ll look at the forces that are parallel to the plane. With the object being on a smooth surface and no external forces, the only force that’s acting parallel to the plane is the parallel component of the weight. So 𝑚𝑔 sin 𝜃 is the net force parallel to the plane, and that is equal to 𝑚𝑎. Since there’s an 𝑚 on both sides of our equation, they can cancel each other out, leaving us with an equation for the acceleration of our box of 𝑔 sin 𝜃. As our slope goes to 90 degrees, sin 𝜃 becomes closer and closer to one. And the acceleration of the object approaches free fall or 𝑔.

But what if the box is on a rough surface instead of a smooth surface? The force of friction will be parallel to the plane and, in this case, will be going up the incline as the box will be trying to slide down the incline. Because the force of friction is parallel to the plane, we must include it in the net force for the forces that are parallel. Our Newton’s second law equation becomes the weight parallel to the plane minus the force of friction is equal to 𝑚𝑎. We have chosen to make down the incline a positive number as that is the motion of the box, thereby making up the incline — in this case, the force of friction — to be negative.

To isolate the friction force, we must add the force of friction to both sides of the equation, thereby canceling out the force of friction on the left side. Then we must subtract 𝑚𝑎 from both sides of the equation, thereby canceling out 𝑚𝑎 on the right side, leaving us with an expression for the force of friction of 𝑚𝑔 sin 𝜃 minus 𝑚𝑎. Now that we have learned how to analyze weight, friction, and normal reaction forces for objects on sloped surfaces, let’s do a few example problems.

A point mass 𝑃 is on a slope, as shown in the diagram. The weight of the mass, 𝑊, and the normal reaction force, 𝑅, act on the point mass. What is the relationship between 𝑅 and 𝑊 if the angle of the slope above the horizontal is zero? What is the magnitude of 𝑅 if the angle of the slope above the horizontal is 90 degrees?

In the first question, it asked us to find the relationship between 𝑅 and 𝑊, where 𝑅 is the normal reaction force and 𝑊 is the weight of the object when the angle of the slope above the horizontal is zero. Let’s redraw our diagram where we have an angle of zero. If the angle is zero degrees, that implies that we have a horizontal surface. Applying Newton’s second law to the situation using only the vertical forces as those are the only forces acting on our point, we get that 𝑅, the normal reaction force, minus 𝑊, the weight, is equal to zero, as our point mass is not accelerating up and down.

We use the common convention that up is positive, that’s why that 𝑅 has a positive value, and that down is negative, which is why the weight 𝑊 has a negative value. This means the magnitude of the normal reaction force is equal to the magnitude of the weight. Therefore, the relationship between 𝑅 and 𝑊 is that they are equal. For the second question, we are asked to find the magnitude of 𝑅 if the angle of the slope above the horizontal is 90 degrees. Let’s once again draw our diagram with an angle of 90 degrees this time.

To solve for this question, let’s find a relationship for 𝑅 in terms of 𝜃. If we align our coordinate system to be parallel and perpendicular to the plane, then we must break down our weight vector into parallel and perpendicular components. We can see the normal reaction force will have the same magnitude as the perpendicular component of the weight. To find the perpendicular component of the weight, we can use trigonometry as the perpendicular component is the adjacent side of a right triangle to the angle 𝜃 and the weight 𝑊 is the hypotenuse of the triangle. So we can use the equation the cos of the angle 𝜃 is equal to the adjacent side of a triangle divided by the hypotenuse of the triangle. Plugging in our values, we have cos 𝜃 is equal to the perpendicular component of the weight divided by the weight.

To isolate the perpendicular component of the weight, we multiply both sides by 𝑊, canceling out the 𝑊 on the right side of the equation, leaving us with the weight of the object times the cos of the angle 𝜃 is equal to the perpendicular component of the weight. We said earlier that the magnitude of 𝑅 will be equal to the magnitude of the perpendicular component of the weight, so we can replace the perpendicular component with 𝑅. We should also recall that weight is 𝑚𝑔, so we could replace the weight 𝑊 with 𝑚𝑔. So the expression for the normal reaction force is 𝑚𝑔 cos 𝜃. The cos of 90 degrees is zero, which means that the normal reaction force of point mass 𝑃 when the angle of the slope is 90 degrees is zero. The magnitude of 𝑅 as the angle of the slope above the horizontal becomes 90 degrees is zero.

An object has a mass 𝑚 is equal to 1.3 kilograms. The object is on a smooth slope with an angle 𝜃 equals 33 degrees, as shown in the diagram. What is the acceleration 𝑎 of the object parallel to the surface of the slope?

Let’s begin to solve the problem by drawing in the forces on our object. The weight of the object pulls straight down, and the normal reaction force acts perpendicularly to the surface. Since the slope is smooth, we do not need to worry about friction. The question asks us for the acceleration of the object parallel to the surface of the slope. Therefore, we need to look at any forces that also happen to be parallel. Only the weight can have a component that is parallel as a normal reaction force is perpendicular to the surface. Let’s break down the weight into its components. The only component that we care about is the parallel component of the weight.

Looking at the right triangle that we drew between the weight and its components, we can see that the parallel component of the weight is the opposite side of the triangle to the angle 𝜃 and the weight 𝑊 is the hypotenuse. We can use trigonometry to determine the parallel component of the weight using sin of 𝜃 is equal to the opposite side of the triangle divided by the hypotenuse. Plugging in our variables, we’d have sin 𝜃 is equal to the parallel component divided by the weight. To isolate the parallel component, we multiply both sides of the equation by 𝑊. This cancels out the 𝑊 on the right-hand side. Our expression for the parallel component of the weight becomes 𝑊 sin 𝜃.

We should remember that the weight of an object is equal to the mass of the object times the acceleration due to gravity, so we can replace the weight with 𝑚𝑔. Now we can plug in the values from our problem. We replace mass with 1.3 kilograms, acceleration due to gravity with 9.8 meters per second squared, and the sin of the angle 𝜃 with the sin of 33 degrees, which gives us a value for the parallel component of our weight of 6.94 newtons. How does that help us find the acceleration in the parallel direction?

We should recall that Newton’s second law is a relationship between the net force and the acceleration of an object, 𝐹 net equals 𝑚𝑎. The net force parallel to the surface of the slope, we just found to be 6.94 newtons. And the mass of the object was given to us as 1.3 kilograms. To isolate the acceleration, we divide both sides of the equation by 1.3 kilograms, thereby canceling out 1.3 kilograms on the right-hand side. When we divide our two numbers, we get 5.3 meters per second squared. The acceleration of the object parallel to the surface of the slope is 5.3 meters per second squared.

Key Points

Use Newton’s second law of motion, 𝐹 net equals 𝑚𝑎, to find the normal reaction force on an object on a sloped surface. Break down the weight of an object into parallel and perpendicular components for an object on a sloped surface. When a smooth sloped surface is at 90 degrees above the horizontal, the normal reaction force is zero.