welcome to I electr online here's the second part of the problem we started on the previous video where we're driving a wedge underneath this object trying to lift it up and we're trying to find the force required to be just at the moment where the object will begin to move it's called the pending motion up situation well the minimum force required to get to that point on the previous video we were able to calculate the forces acting on the Block which was the weight R2 and R1 R1 and R2 are reactionary forces which are a vector sum of the normal forces added to the friction forces and we found the values of those two reactionary forces in terms of the weight of the object we also calculated the angle required between the reactionary force and the normal force to account for the friction Force which is caused by the coefficient of static friction being 0.35 the angle was 19.2 n° now what we doing is we're looking at the wedge itself when we look at the wedge we have the force trying to drive the wedge in we have the reactionary Force at the bottom of the wedge relative to the floor and I guess we have to realize that the wedge is being driven over a floor so we'll add a floor there and we have the reaction of force at the top of the wedge which is between the block and the wedge notice that the R2 that we found at the bottom of the block is the same as the R2 that we find on top of the wedge these are equal and opposite reaction forces that are opposite to one another we already found the value of R2 but we don't yet know the value of R3 and we don't know yet the the value of the force which is ultimately what we're looking for so when we draw a diagram or a vector sum of those three forces of course they must add up to zero that's what it will look like now we know the value for R2 so R2 is a known quantity but we don't yet know R3 and we don't not yet know F so before we can calculate those we're going to need those angles looking at R3 we can see that it makes an angle of 19.2 n° relative to the vertical that's this angle right here so this would be the angle Fe which means that this angle here would be 90° minus V this is 90° minus Fe which is 19.2 n° which is equal 2 that would be 70.6 NS 61 but 71° 71 + 29 is 100 so that is correct so that angle is a 7.71 de angle we can also find this angle right here noted that this angle will be 90° minus the sum of these two angles which is 27.29 De which mean that this angle is equal to 62.7 1° 62 that's 99 that's correct so now we know those two angles one more angle would be this angle right here which is 180° minus 62.7 1° and minus 7.71 de which means that angle is of course we need a calculator 180 minus 62.7 and minus 7.71 equal 46.58 de again we're going to use the law of signs to try to find the values for f and R3 we can say that F divided by the sign of the angle directly across F which is this angle right here the S of 46.58 De is equal to R2 divided by the angle directly across which is the S of 7.71 De I should say the S of the angle directly across which is equal to R3 / by the sign of the angle directly across which is the S of 62.7 1° that means that R3 can be found in terms of R2 to be the ratio we have S of 62.7 1° / the S of 7.71 De and not R but F the force required to drive the wedge in in terms of R2 two is going to be o we still need the ratio here R2 that would be the sign of 46.58 De divided by the S of 7071 Dee all right so 62. 71 take the sign of that and divide by 70.7 and take the sign of that equals and then we multiply that times R2 so this would be 0.942 R2 should be an R there we go and since R2 is known so this is equal to 0.942 * 1373 * the weight of the block so times 1.7 oop 1373 equals that would be 1.29 * the weight of the block so R3 the reactionary force on the bottom surface of the wedge will be equal to that and the force required ultimately what we're looking for that would be 46.58 take the sign of that divided by 7.71 take the sign of that equals and that would be equal to 0.76 well 770 R2 which is 0.770 * 1373 * the weight of the block so times 1373 equals and that would be 1.06 times the weight of the block so a little bit more than the weight of the block to lift it up the reason probably why it's greater than the weight of the block is because the coefficient of static friction is relatively High which means it's a lot of force required to overcome the friction as well as well as the weight of the block but that's how we do that I know we did it in two parts because I need a lot of board space but that's how you find the force required to use a wedge to drive a block up in this particular situation and that's how it's done