A man pushes on a piano with mass 180 kg; it slides at constant velocity down a ramp that is inclined at 19.0° above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (b) parallel to the floor.

Video transcript

Everyone in this problem. We have a 145 kg box placed on a frictionless incline. Okay. With an angle of 22° above the horizontal. The box is gonna slide at constant velocity down the incline as you push it. Okay? We're asked to determine the magnitude of the force you apply on the blocks if the force is parallel to the horizontal. Alright, so let's draw a little diagram here, we have an incline At 22°. Okay, we have our block. That is a 145 kilograms. All right, And you have a person who's going to push it parallel to the horizontal and that block is gonna slide down the incline. Alright, so converting this to a free body diagram, whoever middle point here that represents our block. Okay, we have the normal force which always points perpendicular to the surface. We have our weight to always point straight down in the weight which is equal to MG. We have the force that is applied, which is applied. The force actually, and that's it. Okay, we're told that it's frictionless. We don't have to wear a bit of friction force. The next thing we need to do is pick a coordinate system. Okay, now, in this case we can pick the regular coordinate system that goes kind of up and down um like this and and would be the only thing in that coordinate system perfectly that we would have to decompose. Okay, we're gonna choose the tilted coordinate system. Okay, Because I think it's good practice to get used to working with these tilted coordinate systems. Okay, so we're gonna choose this tilted coordinate system where we have X. Here and y. Going kind of up into the brain. Alright, so our normal force and therefore is in our regular coordinate system, that we chose the other two forces, F and w are going to have to be decomposed. Okay, So if we decompose w we have this W Y. We have W. X. And if we decompose F, we're gonna have F. Y. Pointing up towards the normal force, an F. X. And one thing I want to note right off the bat here is that we've chosen to put the force F pointing to the right, It's not clear which direction it actually points it. Okay? We're told that we are pushing the block box parallel to the horizontal, but we don't know if we're kind of pushing upwards or for pushing downwards. Okay, So we've chosen to push the block downwards. We pointed the force to the right, but what you'll see is when we do the calculation, okay, the sign of that force is going to tell us what direction that force actually points in. Okay, So we've assumed it goes to the right, but the sign might tell us otherwise when we get to the calculation. Okay, Alright, so we're asked to determine the magnitude of the force we apply on the box. Okay? Alright, so let's start in the X direction. We know that the sum of the forces in the X direction. According to Newton's law is equal to mass times acceleration. Okay, now we're told that this is constant velocity when we hear constant velocity, that means zero acceleration. Okay, so this is just going to be equal to zero. So some of the forces in the X direction. What are they? Well, we have this component of the force F. In the X. Direction but we have the component of the weight in the X direction. Okay, so we have F. X plus W. X. These are equal to zero. What is F. X. What is the F. The X. Component of F. This is gonna be F coast data. Mhm. And what about the X component of W Well, this is gonna be M. G. Science. This is all equal to zero. So our force F. It's going to be equal to negative MG sign theater divided by coast later. Okay. And if we plug in our numbers we have the weight. M Our story. The mass M is 145 kg. So we get negative 145 kg. Times gravitational constant 9.8 m per second squared. Okay, acceleration due to gravity there And then we have sign over coast that just gives us 10 as we get 10 of 22°. All right. Now, if we work this out, what we get is that the force is equal to negative 574.12. Okay, So because it's negative, it actually means that it's pointing in the opposite direction that we had drawn it. Um And that's okay because in this problem we're just being asked for the magnitude. Okay? So when we look at the magnitude of that force f We find that it is approximately 574 newtons. Okay, go back up to our answer traces. We have answer. A the magnitude of the force we apply on the block Parallel to the horizontal is approximately 574 newtons. Thanks everyone for watching. I hope this video helped see you in the next one.