A small 1.0 g block charged to 75 nC is placed on a 30° inclined ...

Question

A small 1.0 g block charged to 75 nC is placed on a 30° inclined plane. The coefficients of static and kinetic friction are 0.20 and 0.10, respectively. What minimum strength horizontal electric field is needed to keep the block from sliding down the plane?

Accepted Answer

Hello everyone. Let's go through this practice problem. A 20 g nonconducting cuboid lying at rest on a degrees incline has a total charge of eight multiplied by 10 to the power of negative six columns. The friction coefficient between the cuboid and the incline are mu sub S equals 0.15 and mu sub K equals 0.08. If the setup lies in a horizontal electric field, what least electric field strength is required to keep the cuboid from sliding. Option A 8.3 multiplied by 10 to the power of three nuance per Coolum. Option B 7.3 multiplied by 10 to the power of three nuances per Coolum. Option C 1.7 multiplied by 10 to the power of three and nuance per Coolum. And option D 1.6 multiplied by 10 to the power of four new ones per Coolum. Ok. So this problem has quite a bit going on. So we have an inclined plane which has a cuboid on it and it is within some electric field based on the way the problem is written without that electric field, this cube would slide down the incline but we're trying to create a horizontal electric field that will pull the cuboid back and hold it in place. So let's begin by creating a free body diagram for the cube and analyzing the different forces acting on it. First off, there's going to be the force to due to weight acting downwards the MG second off, there's the normal force acting on it from the surface of the ramp. And next, there's the force of friction which is going to oppose its movement as it tends to slide down the ramp. This is a frictional force. And normally we, we might have to consider the two different types of friction, static friction and kinetic friction. And the problem even gives us the coefficients for both types. But since we're aiming to have the cube perfectly still and at rest here, the coefficient of static friction is more relevant. So I'm gonna write this as F sub S the static friction. And lastly, there is a force acting horizontally in the opposite direction from the electric field which I'm calling F sub E. This is the one that we want to quantify. So let's choose a tilted coordinate plane and begin writing our force equations. Let's begin with our forces in the horizontal direction. So F the net force is in the X direction and note that because we're looking for the minimum field strength, the strength that will keep it at rest, we know that the net force has to be zero. So from the diagram, we can see that there is going to be pulling it to the left the horizontal component of the electric force because our coordinate plane is tilted, there are going to be two different components of this electric force. So I'm specifying the horizontal component, there's also going to be the static friction and then in the positive direction somewhat to the right there is the X component of the force to do to the weight. So that's the basic equation. But now let's start expanding it out a little bit into terms and variables that we actually know first off in regards to the electric force, recall that an electric force due to an electric field is equal to the magnitude of the charge of the particle multiplied by the strength of the electric field, which is what we want to find. So in this case, the this term can be written as the first term for the electric force can be rewritten as the negative of the charge multiplied by E. And since it's the horizontal component due to this angle, theta, here we multiply it by the cosine of theta so that it represents the X component and just minus the frictional force. And recall that friction is equal to the normal force multiplied by the coefficient of friction. In this case, the coefficient of static friction and then plus the force due to the weight which we already mentioned is MG. And since it's the X component, we do what we normally do in this case and multiply by the sign of theta. Now let's move on and discuss the force, the forces in the Y component. Once again, the net force is zero if we assume the cube is at rest, so acting upwards is the normal force opposing, that is the Y component of gravity and also compo opposing, it is the Y component of the electric fields force. Just as we did. For the X forces, we can expand this out a bit as N minus the Y component of the gravitational force is just written as MG cosine theta. And the Y component of the electric force can be written as QE multiplied by the sine of theta. And this equation tells us algebra solving algebraically for the normal force N, we find that the normal force is equal to MG cosine theta plus QE sine theta. So now we can combine the two force equations we made by substituting N in for the X equation we found earlier. So doing that, we find that the forces in the X direction can alternatively be written as and and sub substituting all of the normal force stuff. In four N, we find the negative mu subs the coefficient of static friction multiplied by in parentheses, MG cosine theta plus QE sine theta and that of course is then plus MG sign data minus UE cosine theta and that's all equal to zero. So now we have this all written out as one equation and notice that the only thing that we don't have, the only thing that's not given to us is e the thing we're trying to find in the first place. So really, the rest of the steps to this problem are basically just a matter of rewriting this equation algebraically to solve for the electric field strength. E. And that's all we have to do. Of course, it's a pretty messy equation. So you can get a little intimidating. But as long as you know, your algebra rules, this shouldn't be too bad. So first, I'm going to start by distributing out the coefficient of static friction terms the parentheses on the right, on the left hand side of the equation. And then I'm going to start bringing to the right side of the equation. The terms that don't involve the electric field strength E. So for example, subtracting from both sides of the equation MG sine theta. Notice that every single term has a minus sign so they can all be canceled out to be turned into much more neat looking plus signs. And then we can subtract from both sides of the equation. The only term left on the left hand side, not containing E which is the MG coast data term. And don't forget to include the MU subs over there. So QE cosine theta equals MG sine theta minus mu sub smg cosine theta, let's factor things out to simplify further. So on the left hand side, every term has a QE. So QE in the parentheses, Mu subs sine theta plus cosine theta equals, then we factor out the MG or sine theta minus mu sub S cosine theta and then divide both sides of the equation by everything. On the left hand side, that is an E they get E completely on its own. So we have MG sine theta minus mu sub S cosine theta all divided by Q multiplied by mu subs sine theta plus cosine theta. And that's our final formula. And all that's left for us to do is plug in the values that were given to us in the problem to solve for the electric field strength. So that is equal to 20 g or 20 multiplied by 10 to the power of negative 3 kg multiplied by the gravitational acceleration 9.8 m per second squared multiplied by the sign of 25 degrees minus the coefficient of static friction 0. multiplied by the cosine of 25 degrees. All divided by the electric charge of the cuboid, which is eight multiplied by 10, raise the power of negative six columns multiplied by the coefficient of static friction 0.5 multiplied by the sine of 25 degrees plus the cosine of 25 degrees. And if we put all that into our calculator, then we find an electric field strength of approximately 7.3 7. multiplied by 10 race, the power of three New Ins Peron. And so that is our answer to this problem. And if we look at our multiple choice options, we can see that this agrees with option B. So option B is our final answer to this problem. And that is all for this video. I hope this video helped you out and please consider checking out some of our other videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

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A small 1.0 g block charged to 75 nC is placed on a 30° inclined plane. The coefficients of static and kinetic friction are 0.20 and 0.10, respectively. What minimum strength horizontal electric field is needed to keep the block from sliding down the plane?

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