So in this question we have the following situation, we have a block that's being pulled with the force F at an angle of data. So first let's complete our free body diagrams. We got normal force down, that's our first gravity down. Normal force up force of friction to the left. Now we also have to take this F at the angle and we got to break it up into its components are gonna get rid of this and break it up into how much of that is horizontal and how much of that is vertical. And we're going to do that is horizontal. There's gonna be f coast data and vertical is going to be f sign data and what we're gonna end up with is the following two expressions up and down. We're gonna be balanced. So G equals F. N plus at the signing peter and left and right. Let's assume that they're unbalanced. They don't necessarily have to be but at least if we start there you can always go backwards minus force of friction. Okay, so some of the things that we know, you know tha is 20 degrees. You know, force of friction is equal to 0.5. Plant and MGr force of friction. Just of course F. And we also have some formulas that we can use F. G equals M. G. Uh Net equals M. A. And force of friction static is less than or equal to us. And the normal force and force of kinetic friction is equal to UK comes from a force. Okay, so the first thing we have to do is let's determine our force of friction and our force or f coast data. Yeah, so F coast data, he's going to be equal to 0.5 MG and the co sign of 20 degrees. Okay, we plugged that into our calculator, Plugging in 9.8 for G. We get F costa Equals 46044 M. And yeah, so what we need is we need the force of static friction to be less than that in order for our block to be accelerating. So we have two different situations in part A us is 0.6 um UK is 0.4 That .4.5 mm. So let's look at our above situation and figure out the force of friction so we can move this top expression around and say that F. N. Is equal to F. G minus F sign here. So part of that means that our force of static friction is less than or equal to. We can force a friendship in new s times. Normal force, that's .600 times. You said normal forces, F G minus F santa. So it's M. G minus 0.5 MG times a sign. About 20 degrees. Yeah. Okay, so when we plug all this in, we get force of static friction is less than or equal to 4.87 M. Okay, so since the force of static friction is bigger than our F coast data, that means that F coast data is not overcoming static friction. So the acceleration has to be zero because it's not going to be moving it's not going to be able to overcome the static friction. Now if we go to part B, the only thing that's changed here Is that the coefficient of static friction is now only .4. So let's try this out. 2.4 times. M. G minus 0.5 MG. Uh huh. Can you sign of 20 degrees? Yeah. Yeah. And when we plug this in here. Mhm. Okay. We get force of static friction is less than or equal to 3.2496. M. Okay so this is good. Now we've overcome static friction so we gotta switch. So now let's calculate our acceleration using kinetic friction. So we'll start with our equation M. A. Equals F. G. Cost data. So 0.5 MG comes to co sign up 20 degrees minus our force of kinetic friction which in this case will be Our coalition of kinetic friction is only .3 times. Yeah MG -1 MG. Sign 20° and I ran out of room a little bit there at the end. Um But you get the idea so all of the ends actually go away which is really nice. So I am here and there and there and there. So plugging in 9.8 Fergie and solving you get acceleration is equal to 2.17 m per second squared. So in part A we did not overcome static friction, so the acceleration has to be zero because the block isn't going to move, it's going to remain stationary. In part B, it is going to move. So we have to switch over to kinetic friction. When we saw for acceleration, we get 2.17 m/s squared.