welcome to electron line now that we've finished part 1 in the previous video we're now ready to do part 2 we're trying to find the force required to pull the wedge out in the previous video we looked at the forces acting on the block now we're going to look at the forces acting on the wedge again the force to pull the wedge out that's what we're looking for and we know there's going to be friction on the top surface and on the bottom surface and there's also going to be what we call reactionary forces which are the vector sum of the normal force which is normal to the incline and thus and the friction force I notice the friction force will be acting in this direction the normal force acts in this direction and r1 the reaction Air Force is the vector sum of those two forces r3 again there will be a friction force in this direction as you're trying to pull the the wedge out that'll be a friction force in this direction and there'll be a normal force against the surface here so we add those two together and you get the reactionary force since nothing has started moving yet we're at the moment the pending motion down moment we can add the three forces together and that should add up to zero and this is the vector sum of those three forces what we need to do now is find the three angles made by those three forces also notice that in the previous video in part 1 we already figured out the value of r1 in terms of the weight of the block and that'll allow us then to find the relative force and r3 here as well now how do we find those angles well first of all since the coefficient of static friction is 0.35 we can find the angle relative to the normal of each of the reactionary forces of 19.2 9 degrees which is what we find here the normal is perpendicular so here we can see the angle between r3 and the normal here we have the angle of 19.2 9 degrees between the reactionary force r1 and the normal but we have to subtract the 8 degrees because it's at a slant so the angle with the vertical is only 11 point 2 9 degrees there so how do we find the angles well here we have our 9 this is the angle relative to the vertical so to find this angle here we can say that this angle is equal to 90 degrees minus this angle which is 11 point 2 9 degrees which is equal to 88 point seven one degree if you add this together oh not 88 that would be too big how about seventy eight point seven one degrees so that's 89 that's exactly 90 degrees now for the second angle we can use our three we can come up here we could realize that this angle here is going to be 90 degrees minus the angle relative to the vertical which is nineteen point two nine degrees so this becomes equal to 70 point seven one degrees and finally the third angle can be found by taking 180 degrees and subtracting these two angles from it so minus seventy eight point seven one degree and minus 78 point two seven one degrees and that becomes equal to 180 minus 70 point seven one and minus 78 point seven one equals thirty point five eight degrees now we can use the law of sine to find our 1 and F or not our one that our 3 and F what we can say is that F divided by the sine of the angle directly across which is this angle right here thirty point five eight degrees is equal to r1 divided by the angle directly across which is this angle the sine of 70 point seven one degrees which is equal to r3 divided by the angle directly across which is 70 the sine of seventy eight point seven one degrees all right that allows us to find r3 and f r3 is equal to r1 times the sine of seventy eight point seven one degrees divided by the sine of eight point seven one degrees which is equal to seventy eight point seven one take the sign of that divided by seventy point seven one take the sign of that equals that's equal to 1.0 three 9r1 and of course that is one point zero three nine times R one is zero point nine five three times the weight so times point nine five three equals and it's equal to 0.99 times the weight all right we do the same for F the force which is ultimately what we're trying to find the force required to pull the wedge out that will be equal to r1 times the sine of thirty point five eight degrees divided by the sine of 70 point seven one degrees so that would be thirty point five eight take the sine divided by seventy point seven one take the sine of that equals that's equal to zero point five three nine times r1 which is zero point five three nine times zero point nine five three W the weight so it's R one is ninety five point three percent of the weight and so times 0.95 three equals we have this is equal to 0.51 times the weight and so that's the force required to pull the wedge out it's slightly more than half the weight of the block and that will allow us just like the wedge out if you remember right from the previous videos it took a little bit more than the weight of the block to push the wedge in it make sense and that's how we do a problem like that