in this problem we're going to talk about the towing of a car on an inclined plane so let's say we have an inclined plane and the angle of inclination is at 20 degree and a car is being towed up so it is being pulled upwards by a force the force this force is given which is 5000 Newton the mass of the car is 1000 kilogram and it is being towed off or it is being moved up by a constant speed there's no acceleration and assuming the friction force is negligible we need to know now how much force is acting on to this ramp the maximum force the ramp can with the stand is 9,500 Newton so when I'm moving in this car of course well will the ramp collapse or it will survive let's do the the calculation again the first thing we need to do is find out all the forces acting on to the system or in other words draw a Freebody diagram and draw the forces in the two rectangular components X component and the y component so just to make life easier what we're going to do we're just choosing the decline plane is our x axis and perpendicular to that is we call the y axis so this is our X system or x coordinate and this is y coordinate and I take a look at all the forces we're now applying a force this is the applied force okay and the normal force is always perpendicular to the surface so this is the normal force so we given the two forces apply the force and the normal force and there is another force acting on to this one the weight of the car is always downward so this is given as the weight of the car so three forces we have so far though that one is applied force the normal force and the weight which is the downward and now what we're going to do is resolve all the forces into two components you can see that the force is if this force is in one direction the weight is another direction and the normal force is in the different direction but we may be going to resolve the forces such that all the forces will be in X component and will be along the Y component okay so this force F force and the normal force are already in x and y direction the only force that is not either along X or Y Direction is the weight so we're going to resolve this force now as so this is again normal to the inclined plane as this angle is Theta this angle will also be theta so this angle and this angle is exactly equal and why is that because you see this is the 90 degree so if this angle is Theta this angle will be 90 minus theta so this angle will be theta and this component will be postin component mg cosine theta because this component is closer to the angle and remember the rule see see closer to the angle is always cosine component so this is a closer to the angle so this will be mg cosine theta and this will be mg sine theta because it is further from this angle so now we do not need this force now so what we have to look all the forces along the X direction and all the forces along thing two dicks and that's it so okay so now let's take a look at the all the forces that is acting along the x axis along this axis and you can clearly see there are two forces one is F and the other is mg sine theta and these two forces are acting opposite to each other or in the opposite direction so the total force along the x axis has to be equal to zero now do not get confused the total force is zero why is that because the system the car is not accelerating up it is being tarred up with a uniform velocity so there is no acceleration so we know that the F is equal to mass times acceleration if the system is moving with a constant velocity acceleration would be zero that means the total force will be zero okay and now what is the total force the total force is f minus mg sine theta that means this force is now equal to mg sine theta the mass is given which is thousand kilogram G value you know that is 9.8 sine theta theta is 20 degree so this is all numbers now if you plug in what you get is 3 3 5 2 Newton so the rope as it mentioned here the rope can handle five thousand Newton force and a while pulling up the car we are applying only the 3352 Newton so that means the rope will not break it will survive ok the second part is will the so this rope here if it is being pulled by a rope the rope is not break for sure now now see if this ramp will collapse how we going to find out how was force is acting on to this ramp this inclined plane the total force that is acting on to the Ram is this force mg cosine theta or the normal force these two forces are equal and opposite so we need to find out they what is the normal force here so let's do that so if we look at the along the y axis along this axis again the total force is L because the system is not moving along the Y direction so the total force along this axis must be zero so FY is equal to in general sorry for some reason it collapsed let me open it again so this is now n minus mg cosine theta this has to be equal to zero and now n will be equal to mg cosine theta the mass is given G is given cosine theta and if you saw what I get is ninety two hundred and nine Newton force and this ramp can handle ninety five hundred Newton so again it won't collapse our safe collapse so that's how you're going to solve the inclined plane problem okay and again here I'm assuming there is no friction force if there's a friction force you have to take into account the the friction force as well which will be indeed a downward direction if it is moving up again if you have any questions write down your questions in the comment section below and at the end do not forget to like share & subscribe the channel thank you very much