Question Video: Finding the Force That Halves the Acceleration of a Body on an Inclined Plane | Nagwa Question Video: Finding the Force That Halves the Acceleration of a Body on an Inclined Plane | Nagwa

# Question Video: Finding the Force That Halves the Acceleration of a Body on an Inclined Plane Mathematics • Third Year of Secondary School

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A body of mass 205 kg was left to slide down a plane inclined at 45° to the horizontal. A force started acting on the body causing its acceleration to halve. Given that the line of action of the force made an angle of 45° with the line of greatest slope of the plane and that they lie in the same vertical plane, find the magnitude of this force. Consider the acceleration due to gravity to be 9.8 m/s².

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### Video Transcript

A body of mass 205 kilograms was left to slide down a plane inclined at 45 degrees to the horizontal. A force started acting on the body causing its acceleration to halve. Given that the line of action of the force made an angle of 45 degrees with the line of greatest slope of the plane and that they lie in the same vertical plane, find the magnitude of this force. Consider the acceleration due to gravity to be 9.8 meters per square second.

Before we can answer this question, we’re going to begin by drawing a sketch of the scenario. This sketch doesn’t need to be to scale, but it should be roughly in proportion so we can accurately model what’s happening. We have a body with a mass of 205 kilograms sliding down a plane inclined at 45 degrees to the horizontal. Since the body has a mass of 205 kilograms, we can say it exerts a downward force on the plane and this force is equal to mass times acceleration due to gravity. If we let acceleration due to gravity to be 𝑔 — and we’ll eventually use 9.8 — we see that the downwards force of the body on the plane is 205𝑔 newtons.

We can also define its acceleration in the first part of its journey to be equal to 𝑎 meters per square second. Then, a force acts on the body causing its acceleration to halve. Now, we’re not going to model this just yet. Instead, we’re going to use what we have so far to calculate the acceleration during the first part of the journey. The formula we’re going to use to do this is force equals mass times acceleration. Now, we’ve modeled the acceleration as being parallel to the plane, so we need to work out the components of the forces which are also parallel to the plane.

At this stage, there’s only one force in this diagram. And so, we add a right-angled triangle. And we see that we need to work out the value of 𝑥; this is the component of the weight force that’s parallel to the plane. The included angle in this triangle is 45 degrees. Now, we’re looking to find the opposite, and we know the hypotenuse is 205𝑔. We can, therefore, use the sine ratio. sin 𝜃 is opposite over hypotenuse. So, in this case, sin 45 is 𝑥 over 205𝑔. If we multiply through by 205𝑔, we find 𝑥 is equal to 205𝑔 times sin of 45, which is 2009 root two over two.

Now that we found the component of the weight force that’s parallel to the plane, let’s substitute what we know into the formula 𝐅 equals 𝑚𝑎. This gives us 2009 root two over two equals mass times acceleration, so 205 times 𝑎. We’ll solve for 𝑎 to calculate the acceleration during the first part of the journey by dividing through by 205. And when we do, we find 𝑎 is 49 root two over 10 or 49 root two over 10 meters per square second.

Now that we’ve calculated the acceleration during the first part of the journey, let’s look at the second part of the journey. We’re told that a force acts on the body that causes the acceleration to halve. If we define the acceleration in the second part of the journey to be 𝑎 sub two, we can say it’s half of 𝑎. It’s half of 49 root two over 10 or 49 root two over 20 meters per square second.

Let’s now go back to our diagram and add in the new force. We’re told that it makes an angle of 45 degrees with the line of greatest slope of the plane. In other words, it acts directly opposite to the weight. And so, how do we calculate the value of 𝐅? Well, once again, we’re going to use the formula 𝐅 equals 𝑚𝑎. We know the mass and the acceleration of the body at this point, but what is the overall force?

We still have the force of the weight that acts parallel to the plane. But if we add a right-angled triangle to the force 𝐅, we see we need to calculate the component of this force that’s also parallel to the plane. This is the adjacent side in the triangle, and we know the hypotenuse. Defining the adjacent side in this triangle to be equal to 𝑦, then we can link 𝑦 and 𝐅 using the cosine ratio. We get cos of 45 equals 𝑦 over 𝐅. And then, we multiply by 𝐅. And we see 𝑦 is 𝐅 times cos of 45 or root two over two 𝐅.

Let’s clear some space and complete the formula 𝐅 equals 𝑚𝑎. If we consider the positive direction to be the direction that the body is moving, that is down the plane, then we know we have a positive force of 2009 root two over two newtons. We’re going to subtract the component of the force 𝐅 that is parallel to the plane. And we’re subtracting because this is acting in the opposite direction. So, the overall force is 2009 root two over two minus root two over two 𝐅. This is equal to mass times acceleration. And now, we’re going to use the new acceleration, the halved acceleration; that’s 49 root two over 20.

And so, we have an equation for 𝐅. To solve, we can begin by dividing through by the square root of two. We might also choose to multiply both sides of the equation by two, just to get rid of these fractions. And so, our equation becomes 2009 minus 𝐅 equals 205 times 49 over 10. 205 times 49 over 10 is 1004.5. So, let’s add 𝐅 to both sides and then subtract 1004.5. When we do, we get a value for 𝐅 as being 2009 minus 1004.5, which is 1004.5. And so, we can say that 𝐅 is equal to 1004.5 newtons.

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