Question Video: Studying Motion of a Body on a Smooth Inclined Plane under the Action of a Given Force Mathematics

A body of mass π‘š was placed on a smooth plane inclined at an angle πœƒ to the horizontal, where tan πœƒ = 3/4. A force of magnitude 77 kg-wt was acting on the body along the line of greatest slope up the plane. Given that this force caused the body to start moving up the plane and that it covered 196 cm in 2 seconds, find the mass of the body π‘š. Take 𝑔 = 9.8 m/sΒ².

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Video Transcript

A body of mass π‘š was placed on a smooth plane inclined at an angle πœƒ to the horizontal, where tan πœƒ equals three-quarters. A force of magnitude 77 kilogram weight was acting on the body along the line of greatest slope up the plane. Given that this force caused the body to start moving up the plane and that it covered 196 centimeters in two seconds, find the mass of the body π‘š. Take 𝑔 to be equal to 9.8 meters per square second.

Before we do any calculations here, we’re simply going to begin by drawing a diagram of the scenario. We have a body with mass π‘š placed on a plane inclined at an angle to the horizontal. Since the body has a mass π‘š, we know that the downwards force it exerts on the plane is mass times gravity. Let’s call that π‘šπ‘” for now. It’s inclined at an angle πœƒ to the horizontal, where tan πœƒ is equal to three-quarters. Now what we’re not going to do with this equation is solve for πœƒ. Instead, we’ll use this later on to work out the exact value of sin πœƒ. We’ll deal with that in a moment.

We then have a force acting up the line of greatest slope of the plane. It’s 77 kilogram weight, as shown. The plane is smooth, so there are no frictional forces acting on the body. And this means the only other force we’re interested in is the normal reaction of the plane on the body, as shown. And so, we’re going to use the equation 𝐹 equals π‘šπ‘Ž; force is equal to mass times acceleration. And we’re going to resolve forces parallel to the plane. We know that the body starts to move up the plane. So, we’re going to assume acceleration is positive in this direction.

And so, let’s find the sum of the forces acting in this direction. We have 77 kilogram weight acting up the plane. Now, we’re going to work in newtons here. And we recall that one kilogram weight is equal to 9.8 newtons. Now, the reason we’re converting to newtons is because we’re measuring the weight of the body, mass times gravity, in newtons. And we need to make sure all our forces are in the same unit. This means to find the number of newtons in 77 kilogram weight, we multiply 77 by 9.8 to give us 754.6 newtons. So, we have a force of 754.6 newtons acting up and parallel to the plane.

But what about the downward force of the body on the plane? This acts in neither a parallel or perpendicular direction. And so, we add a right-angled triangle. And we see that the other two sides in this right-angled triangle are the components of this weight, which are parallel and perpendicular to the plane. The included angle is πœƒ. And we’ll label the side that we’re trying to find which is the side of the triangle parallel to the plane π‘₯ or π‘₯ newtons. We’re looking to find the opposite and we have the hypotenuse. So, we can say that since sin πœƒ is opposite over hypotenuse, sin πœƒ here must be π‘₯ over π‘šπ‘”, meaning π‘₯ must be equal to π‘šπ‘” times sin πœƒ.

But what is sin πœƒ? Well, to find out, we use the fact that tan πœƒ is three-quarters. We can draw a right-angled triangle with an included angle of πœƒ. And since tan πœƒ is opposite over adjacent, we know the opposite side in this triangle is three units and the adjacent is four units. By using the Pythagorean theorem or recognizing that we have a Pythagorean triple, we see the hypotenuse in this triangle is five units. sin πœƒ is opposite over hypotenuse. So, for our value of πœƒ, sin πœƒ is three over five or three-fifths. So, we find that π‘₯ is equal to π‘šπ‘” times three-fifths. We simplify this to three-fifths π‘šπ‘” and recognize that it’s acting in the opposite direction to the force pushing the body up the slope. So, we subtract it. The overall force acting on the body then is 754.6 minus three-fifths π‘šπ‘”. This is equal to π‘šπ‘Ž, mass times acceleration.

But what is the acceleration? To calculate the acceleration, we go to the second to last sentence in this question. This tells us that the body covers 196 centimeters in two seconds. We’re working in meters per square second, so we see that it covers 1.96 meters in two seconds. Its starting velocity, 𝑒, is zero and we’re trying to calculate its acceleration, π‘Ž. So, let’s use the formula that links these four variables: 𝑠 equals 𝑒𝑑 plus a half π‘Žπ‘‘ squared. Substituting our values in and we get 1.96 equals zero plus a half times π‘Ž times π‘Ž squared. 1.96 is, therefore, equal to two π‘Ž. And if we divide through by two, we find the acceleration of the body to be equal to 0.98 meters per square second.

We’re going to replace π‘Ž in our original equation with 0.98. And when we do, our equation becomes 754.6 minus three-fifths π‘šπ‘” equals 0.98π‘š. We now have an equation in π‘š, so let’s solve. We begin by adding three-fifths π‘šπ‘” to both sides. Then we factor by π‘š on the right-hand side and we get π‘š times 0.98 plus three-fifths 𝑔. But of course, we were told to take 𝑔 to be equal to 9.8. So, this right-hand side becomes π‘š times 0.98 plus three-fifths times 9.8. To solve for π‘š, we divide through by 0.98 plus three-fifths times 9.8 and that gives us a value of 110. We were initially working in kilogram weight and we converted to newtons. So, we want our mass to be in kilograms.

The mass of the body π‘š is 110 kilograms.

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