Question Video: Finding the Acceleration of a Body in Motion on a Rough Inclined Plane Mathematics

A body was placed on the top of a rough inclined plane of length 259 cm and height 84 cm. The body started sliding down the plane. Given that the coefficient of friction was 0.29, find the acceleration of the body. Take ๐‘” = 9.8 m/sยฒ.

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Video Transcript

A body was placed on the top of a rough inclined plane of length 259 centimeters and height 84 centimeters. The body started sliding down the plane. Given that the coefficient of friction was 0.29, find the acceleration of the body. Take ๐‘” to equal 9.8 meters per second squared.

Okay, so letโ€™s say that this is our rough inclined plane. And weโ€™re told that it has a height of 84 centimeters and a length of 259 centimeters. And we have this body that weโ€™re told starts out at the top of the incline and then starts sliding down. Knowing that the coefficient of friction between the body and the incline, weโ€™ll call it ๐œ‡, equals 0.29, we want to solve for the acceleration of this body.

To begin doing this, we can recall Newtonโ€™s second law of motion, which says that the net force acting on some object or some body is equal to that bodyโ€™s mass times its acceleration. Returning to our sketch, if we put in a positive ๐‘ฅ- and positive ๐‘ฆ-direction to our diagram, we can say that Newtonโ€™s second law applies in both of these dimensions. In other words, if we consider all the forces acting on our body in what weโ€™ve called the ๐‘ฅ-direction, then the sum of those forces will equal the bodyโ€™s mass times its acceleration in this direction. Thatโ€™s just what we want to solve for. So letโ€™s consider the forces acting on our body as it slides down the plane.

First off, we know thereโ€™s the force of gravity, ๐‘š times ๐‘”. We know thereโ€™s also a normal or reaction force, weโ€™ll call it ๐‘…, acting on our body perpendicular to the plane. And lastly, thereโ€™s a friction force, weโ€™ll call it ๐น. We know this force acts up the plane because our body is sliding down this surface and friction always opposes the direction of motion. So these are our three forces. And if we consider specifically the ๐‘ฅ-components of these forces, then we see that the weight force, ๐‘š times ๐‘”, has an ๐‘ฅ-component to it and that the frictional force is entirely in this dimension.

Now, in our sketch, if we were to call this angle here ๐œƒ so that ๐œƒ defines the angle of inclination, then this angle here in this right triangle would also be ๐œƒ. This means that the ๐‘ฅ-component of the weight force would be ๐‘š times ๐‘” times the sine of this angle weโ€™ve called ๐œƒ. And from that we would then subtract the frictional force ๐น, which gives us the mass of our object multiplied by its acceleration in the ๐‘ฅ-direction. Recall that itโ€™s the bodyโ€™s acceleration along the direction of the plane, what weโ€™ve called ๐‘Ž sub ๐‘ฅ, that we want to solve for. To help us do that, letโ€™s recall that, in general, the frictional force for an object in motion, weโ€™ve called it ๐น, is equal to whatโ€™s called the coefficient of friction ๐œ‡ multiplied by the reaction force acting on that object.

In this instance, weโ€™re given ๐œ‡, 0.29. But to figure out ๐‘…, it will be helpful to take a look at our diagram. Letโ€™s consider that this reaction force ๐‘… is entirely in what weโ€™ve called the positive ๐‘ฆ-direction. We know also that our body doesnโ€™t accelerate in this dimension as it slides down the plane. This tells us that the ๐‘ฆ-dimension component of the weight force, highlighted here, must be equal in magnitude but opposite in direction to the reaction force ๐‘…. That magnitude will be equal to the weight force ๐‘š times ๐‘” times the cosine of the angle weโ€™ve called ๐œƒ.

All this means that we can take the coefficient of friction ๐œ‡, multiply it by ๐‘š times ๐‘” times the cos of ๐œƒ, and substitute it in for the friction force ๐น. Because, as weโ€™ve seen, ๐‘š๐‘” times the cos of ๐œƒ is equal in magnitude to our reaction force. That gives us this equation for the forces in the ๐‘ฅ-direction of our diagram. And notice something interesting here. The mass of our body ๐‘š appears in all three terms. Since this mass is assumed to be greater than zero, we can divide both sides by this factor and cancel it out. If we then factor out the acceleration due to gravity from both terms on the left-hand side, we get this expression. And letโ€™s recall that weโ€™re given ๐‘” as well as ๐œ‡. So all we need to do to solve for ๐‘Ž sub ๐‘ฅ is substitute in for the sin and cos of ๐œƒ.

Looking at our sketch, we can say that the sine of the angle weโ€™ve called ๐œƒ is equal to the opposite side that has a height of 84 centimeters divided by the hypotenuse with a length of 259 centimeters. And then to solve for the cosine of this angle, we use the Pythagorean theorem to state that the length of this side of our right triangle equals the square root of 259 squared minus 84 squared in units of centimeters. That turns out to be equal to 245 centimeters, which means that the cos of ๐œƒ is equal to 245 divided by 259.

If we then substitute in our values for ๐œ‡ and ๐‘” without their units, when we calculate this expression, we find a result of 0.49. This result has units of meters per second squared. And so we say that the acceleration of this descending body is 0.49 meters per second squared.

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