Question Video: Finding the Acceleration of and the Normal Reaction to the Motion of a Body on a Smooth Horizontal Plane Where an Inclined Force Acts on It Mathematics

A body of mass 37 kg was placed on a smooth horizontal surface where a force of 72 N was acting on the body such that its line of action made an angle of 60ยฐ downward from the vertical. Determine the acceleration of the body ๐‘Ž and the magnitude of the normal reaction ๐‘…. Round your answers to the nearest two decimal places.

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Video Transcript

A body of mass 37 kilograms was placed on a smooth horizontal surface where a force of 72 newtons was acting on the body such that its line of action made an angle of 60 degrees downward from the vertical. Determine the acceleration of the body ๐‘Ž and the magnitude of the normal reaction ๐‘…. Round your answers to the nearest two decimal places.

Letโ€™s begin by drawing a free body diagram showing all the information from this question. Here, we have a body resting on a smooth horizontal surface. Now the fact that itโ€™s smooth means that there are no frictional forces that we need to consider between the body and the surface itself. Next, weโ€™re told the body has a mass of 37 kilograms. Using Newtonโ€™s second law of motion, which says that force is proportional to the acceleration of the body โ€” in other words, force is equal to mass times acceleration โ€” we can say that the downward force of the weight of the body, which is its mass times acceleration due to gravity, is 37๐‘”. And we generally take ๐‘” to be equal to 9.8 meters per square second.

Next, letโ€™s consider the external force that acts on the body. The force has a magnitude of 72 newtons, and it acts at an angle of 60 degrees downward from the vertical. Weโ€™ve modeled the downwards force of the weight to be acting vertically downward. So, the angle between the two forces we currently have is, therefore, equal to 60 degrees. Then by Newtonโ€™s third law of motion, since the body itself is exerting a downwards force on the horizontal surface, the surface itself exerts an equal and opposite force on the body. Weโ€™ll label that ๐‘…, although itโ€™s sometimes called ๐‘› for the normal reaction.

So, we now have all of our forces, and weโ€™re looking to find the acceleration of the body ๐‘Ž. Weโ€™ll make an assumption that due to the direction in which the force of 72 newtons acts, the body will accelerate towards the right. And weโ€™re going to define that as in the question to be ๐‘Ž meters per square second. Then weโ€™ll once again use the equation overall force or net force is equal to mass times acceleration. Now, since the 72-newton force acts at an angle, weโ€™re going to resolve it into the component of this force that acts parallel to the horizontal surface, in other words, that acts in the same direction as the acceleration. And to do so, weโ€™ll use the fact that the vertical and horizontal meet at an angle of 90 degrees, and 90 minus 60 is equal to 30.

Then, we can model this as a right-angled triangle. Letโ€™s label the side that weโ€™re trying to find ๐‘ฅ, and we know the hypotenuse is equal to 72 newtons. And this allows us to link these two measurements using the cosine ratio. Cos of ๐œƒ is equal to adjacent over hypotenuse. In this case, cos of 30 is equal to ๐‘ฅ over 72. And then ๐‘ฅ is equal to 72 times cos 30. Now, in fact, we know cos of 30 degrees is equal to root three over two. So, we can write this as 72 times root three over two, which is 36 root three or 36 root three newtons. And so, we calculated the component of our 72-newton force that acts in the same direction as the acceleration.

Then, since the surface is smooth, weโ€™re assuming there are no other frictional forces, so the net force acting on the body in the horizontal direction is 36 root three newtons. Substituting this into the formula ๐น equals ๐‘š๐‘Ž, where the mass of the object is 37 kilograms, and we get 36 root three equals 37๐‘Ž. We can now divide through by 37, and we find that the acceleration is 36 root three over 37, which is 1.685 and so on or 1.69 meters per square second correct to the nearest two decimal places.

We now have the acceleration, but there is another part to this question. Itโ€™s asking us to find the magnitude of the normal reaction ๐‘…. To do so, we make an assumption that in the vertical direction the object is in equilibrium. If this is true, the net sum of the forces acting in this direction will be equal to zero. Our normal reaction force ๐‘… acts upwards and then, in fact, we have two forces acting vertically downwards. Thereโ€™s the downwards force of the weight 37๐‘”, but thereโ€™s also the component of the 72-newton force that acts in this direction. Letโ€™s call that ๐‘ฆ.

Once again, using right triangle trigonometry or we could even use the Pythagorean theorem, we can form an expression for ๐‘ฆ. We get sin of 30 this time equals ๐‘ฆ over 72, which means ๐‘ฆ is equal to 72 sin 30 or 36 newtons. And so, we have the magnitudes of our two downwards forces. We have 37๐‘” newtons, which is the downwards force of the weight, and 36 newtons, which is the downwards component of the 72-newton force. If we take upwards to be positive here, we know that the sum of all the forces in this direction is zero. So, thatโ€™s ๐‘… minus 37๐‘” minus 36 equals zero. And adding 37๐‘” and 36 to both sides, we get ๐‘… equals 37๐‘” plus 36. Since ๐‘” is 9.8 meters per square second, this becomes 37 times 9.8 plus 36 which is 398.6, or 398.60 newtons. So, ๐‘Ž is equal to 1.69 meters per square second and ๐‘… is equal to 398.60 newtons.

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