Question Video: Analysis of the Equilibrium of a Body Resting on a Rough Horizontal Plane Attached to an Inclined String Causing the Body to Be on the Point of Moving

A body weighing 63 newtons rests on a rough horizontal plane. The tangent of the angle of static friction between the body and the plane is 1/8. The body is attached to a light inextensible string which is inclined to the horizontal by an angle whose sin is 12/15. Given that the tension in the rope causes the body to be on the point of moving, find the magnitude of the tension ๐‘‡ and the magnitude of the static friction ๐น.

09:49

Video Transcript

A body weighing 63 newtons rests on a rough horizontal plane. The tangent of the angle of static friction between the body and the plane is one-eighth. The body is attached to a light inextensible string which is inclined to the horizontal by an angle whose sin is twelve fifteenths. Given that the tension in the rope causes the body to be on the point of moving, find the magnitude of the tension ๐‘‡ and the magnitude of the static friction ๐น.

Okay, before we get started on our solution, letโ€™s record some of this information given. Weโ€™re told that weโ€™re considering a body that has a weight of 63 newtons; weโ€™ll call that weight ๐‘Š. And then weโ€™re told about this quantity called the angle of static friction, and specifically weโ€™re told the tangent of that angle. That angle of static friction refers to a scenario where we have a body at rest on a surface. And then, bit by bit, we incline that surface until eventually at a steep enough angle the body just starts to slide. At that moment, the angle that our inclined plane makes with the horizontal is called the angle of static friction.

In our case, weโ€™re told the tangent of that angle. And itโ€™s helpful to note that the tangent of the angle of static friction is always equal to the coefficient of static friction. All this means that, for our particular body on our particular horizontal plane, the coefficient of static friction is one-eighth. Along with all this, weโ€™re told that our body is attached to a light inextensible string. In other words, weโ€™re saying that the string doesnโ€™t stretch at all, and then it has effectively no mass. If we sketch in this string, weโ€™re told that itโ€™s inclined to the horizontal at an angle whose sin is this ratio twelve fifteenths.

In other words, if we were to make a right triangle where the hypotenuse is represented by the string, then the sin of this interior angle here is twelve fifteenths. Since the sine of that angle equals the ratio of this side length to the hypotenuse length, we can say that the vertical side length of this triangle weโ€™ve drawn is 12 units, while the hypotenuse is 15. Based on all this, as weโ€™ve seen, we want to solve for the tension ๐‘‡ in the string and the magnitude of static friction ๐น between our body and the horizontal plane.

Clearing space to work on screen, as we said, itโ€™s the tension ๐‘‡ and static friction force ๐น that we want to find. Starting out solving for the tension ๐‘‡, weโ€™re told that when the string pulls on our body, it makes it so the body is on the point of moving. Since the string pulls both vertically upward and horizontally to the right, the fact that our body is on the point of motion could mean different things. On the one hand, it could mean that our body is about to move vertically upward off of the surface. On the other, it could mean that our body is about to move to the right, that is, that the force of static friction keeping it in place is overcome. And then, technically, thereโ€™s even a third option as well, which is that both of these types of motion will happen at the same instant.

To get more clarity on the forces acting on our body, letโ€™s draw a bit of an expanded sketch. Viewed this way, in addition to the tension force thatโ€™s acting on our body, we knew that thereโ€™s a weight force, itโ€™s mass times the acceleration due to gravity, that acts on it, where here our arrows are not drawn to scale. And so long as the body stays in place, thereโ€™s also a static friction force ๐น acting to the left. Lastly, there may or may not be a reaction force ๐‘… acting on our body. That will depend on how the vertical component of our tension force compares with a downward-acting weight force. These are the forces that act or potentially act on our body while itโ€™s at rest.

Now, as we mentioned, under this tension force ๐‘‡, our body is just about to move. And it could be that that motion is purely in the vertical direction or purely in the horizontal direction or in a combination of the two. We can figure out with certainty which of these three cases it is, but only if we start out by making an assumption. Letโ€™s assume that, under the force of this tension ๐‘‡, our body is about to move but in the horizontal direction only. If thatโ€™s true, it means that all of the horizontal forces acting on the body balance out. When it comes to those forces, we see that thereโ€™s a horizontal component of our tension force and then the oppositely directed force of static friction.

Say that we set up the convention that forces acting to the right are positive. And letโ€™s say further that we give a name to this interior angle of our right triangle. Letโ€™s call this angle ๐œ™. Because thereโ€™s a tension force ๐‘‡ along the string and that the string is an angle ๐œ™ above the horizontal, we can say the horizontal component of ๐‘‡ is ๐‘‡ times the cos of ๐œ™. If we subtract from this the frictional force ๐น, which as we saw acts in the opposite direction, weโ€™ll get zero. Thatโ€™s what it means for these forces to be balanced. We can write therefore that ๐‘‡ times the cos of ๐œ™ equals ๐น. And now letโ€™s consider what this frictional force is.

When this frictional force is static, itโ€™s often represented as ๐น sub ๐‘ . And itโ€™s equal to the coefficient of static friction multiplied by the reaction force experienced by a body. In many scenarios, the reaction force ๐‘… is equal and opposite to the weight force ๐‘š times ๐‘” of a given body. In our scenario though, thatโ€™s not the case. And thatโ€™s because we have this vertical component of the tension force. That vertical component equals ๐‘‡ times the sin of ๐œ™. And if we decided that in the vertical direction forces directed upward were positive, then we could say that the reaction force ๐‘… plus ๐‘‡ times the sin of ๐œ™ minus ๐‘š times ๐‘” equals zero. This implies that the reaction force ๐‘… equals ๐‘š times ๐‘” minus ๐‘‡ sin ๐œ™.

Now, if this vertical component of our tension force was equal and opposite to the weight force, then we see that our reaction force ๐‘… would be zero. But continuing on with our assumption that our body is about to move only horizontally, we can say that the reaction force ๐‘… is not zero. And as we mentioned, later on weโ€™ll get a chance to test whether our assumption is true. For now though, we have an expression to use for the reaction force ๐‘… in our equation for the static friction force. This all means that we can rewrite that force ๐น as ๐œ‡ times ๐‘  times the quantity ๐‘š times ๐‘” minus ๐‘‡ sin ๐œ™.

Recall that, in our problem statement, weโ€™re given a value for ๐œ‡ sub ๐‘ ; itโ€™s one-eighth. And we can also note that ๐‘š times ๐‘”, the mass of our body multiplied by the acceleration due to gravity, is equal to its weight, which we have a variable for and a value for. Clearing a bit of space again, hereโ€™s where we now stand. Under our assumption about our bodyโ€™s motion, ๐‘‡, the tension in the string, multiplied by the cos of ๐œ™ equals ๐œ‡ sub ๐‘ , which is a known value, multiplied by the quantity of ๐‘Š, which is also known, minus ๐‘‡ times the sin of ๐œ™.

We want to solve for the tension ๐‘‡. So letโ€™s first multiply through by ๐œ‡ sub ๐‘  on the right-hand side. And then letโ€™s add ๐œ‡ sub ๐‘  times ๐‘‡ times the sin of ๐œ™ to both sides of the equation. When we do that and we factor out the tension ๐‘‡ on the left, the next thing we can do is divide both sides by the cos of ๐œ™ plus ๐œ‡ sub ๐‘  times the sin of ๐œ™. And if we then remove what cancels out on the left-hand side, we now have an equation where ๐‘‡ is the subject. Since weโ€™re given the values for ๐œ‡ sub ๐‘  and ๐‘Š, what remains is to solve for the values of the cos and the sin of this angle ๐œ™.

Looking back at our diagram and the right triangle weโ€™ve drawn, we recall that actually weโ€™ve already been given the value of the sin of this angle weโ€™ve called ๐œ™. Itโ€™s equal to the ratio of the opposite side length of the triangle to the hypotenuse length. In other words, the sin of ๐œ™ is twelve fifteenths. When it comes to the cos of ๐œ™, that length is equal to this side length of our triangle. Because this is a right triangle and we know the other two side lengths, we can use the Pythagorean theorem to solve for this one. If we call the length of this side of our triangle ๐‘™, then we can say that ๐‘™ equals the square root of 15 squared minus 12 squared. This comes out to exactly nine. So ๐‘™ equals nine and the cos of ๐œ™ then is equal to ๐‘™ over 15 or, in other words, nine fifteenths.

Weโ€™re now ready to substitute all this information in to the right side of our equation. Leaving out units, ๐‘‡ is equal to this fraction. And if we enter this expression on our calculator, we get 11.25. Since the tension is a force, we know that the units of this are newtons. And so the answer to the first part of our question is that the tension ๐‘‡ in the string equals 11.25 newtons. We can now move on to part two, where we calculate the static frictional force ๐น. We saw earlier that that force is equal in magnitude to ๐‘‡ times the cos of ๐œ™. Thatโ€™s equal then to 11.25 times nine fifteenths with units of newtons. This equals 6.75 newtons.

Weโ€™ve now answered both parts of our question, but thereโ€™s still the outstanding issue of if our assumption that we started out with was correct. Before we knew what the magnitude of the tension ๐‘‡ is, we saw that the body couldโ€™ve been about to be in motion in the vertical direction only, the horizontal direction only, or the vertical and the horizontal together. Now that we do know the tension ๐‘‡ in our string, letโ€™s consider what effect this would have on the vertical motion of this body. As we saw, the vertical component of ๐‘‡ is ๐‘‡ times the sin of ๐œ™. That equals 11.25 times twelve fifteenths newtons. This comes out to exactly nine newtons.

So then, when our body is just about to start moving in the horizontal direction, the vertical component of that force is nine newtons. And this is opposed by the weight force of 63 newtons. The fact that the weight force is the greater of these two values confirms the assumption we made. Under this tension force, the body begins to move in the horizontal direction before it begins to move vertically. Our initial assumption then was valid. And so as our final answers, we can confidently say that ๐‘‡ equals 11.25 newtons, while ๐น equals 6.75 newtons.

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