welcome to our lectern line here we have the same setup as we did in the previous two videos except now we're trying to find the force needed to cause the block to slide down what we're trying to do here is remove the wedge have the blocks like that what is the pending motion downward force required to do so so any additional force the block will slide down what's that limit of that force and again what we need to do is we need to take the block and draw the Freebody diagram on the block we have the weight of the block pushing downward since we're going to be applying a force on the wedge trying to pull it this way the friction on between the wedge and the block will cause the block to slide into the wall over here and there'll be a reactionary force over here now the reactionary force will be the sum of the normal force and the friction force caused by the friction between the wall and the block the coefficient of static friction is 0.35 and the angle caused by that between the normal and the reaction force will be 19 point 2 9 degrees we also have friction between the block and the wedge that's friction again the normal force will be perpendicular to the surface which is an angle of 8 degrees with the vertical and then we have a 19 point to 9 degree angle between the normal force and the reaction force but since it's on the other side of the normal we have to subtract the 8 degrees from 19.3 9 and beget an angle of 11 point to 9 degrees relative to the vertical we then take these three forces and we sum them together and since there's no motion there's only pending motion to sum of those forces must add up to zero now what we need to do is find the angles between those forces when we look at r1 we can see that the angle relative to the vertical is 11 point 2 9 degrees which means that's the same over here the angle over here relative to the vertical has to be 11 point 2 9 degrees over here we can see that the reactionary forced r2 makes an angle of nineteen point two nine degrees with the horizontal which means that that's the angle over here and to find this angle over here we need to take 90 degrees and subtract from that the nineteen point two nine degrees which is 70 point seven one degrees and finally to find the third angle right here there we take 180 degrees subtract eleven point two nine degrees and subtract 78 point seven one degree when you add these together that is eighty two subtract from that that means that this would be equal to 98 degrees that's so say 81 yep exactly 98 degrees now that we know the three angles we can use the law of sines to find the values for r1 and r2 that's not yet what we're looking for we're looking for F but we'll do that in part two right now we're going to need to find r1 and r2 which will then be used in the next part bar two to find the actual force needed to pull do my job to do this we take the weight of the block divided by the sine of the angle directly across from that which is this angle right here ninety degrees and we set that equal to r1 divided by the sine of the angle directly across which is this angle let's see here that would be how that crosses over that would be divided by the sine of 70 point seven one degrees and we set that equal to r2 reactionary force two divided by the angle directly across which is the sine of eleven point two nine degrees that allows us to find r1 and r2 in terms of the weight of the block our one will be equal to the weight times the ratio of the sine of 70 point seven one degree divided by the sine of 98 degrees and for that we need a calculator seventy point seven one and we take the sign of that and divided by 98 take the sign of that equals and that gives us value of zero point nine five three times the weight of the block we do the same for our to the weight times the ratio of that would be the sign of eleven point two nine degrees divided by the sine of 98 degrees all right so 11 point two nine take the sign of that and divided by 98 take the sign of that equals n we get zero point one nine eight times the weight so 19.8 percent of the weight so that gives us the values for r1 and r2 let's write that down over here so we can use those for our next part so r1 is equal to zero point eight times the weight and those are necessary then for Part two when we start drawing a Freebody diagram on the wedge itself to find out the force required to pull the wedge out it will be again the pending motion downward as we call it and that's how it's done