00:01
Car of mass 1 .6mg so that's megagrams copper dot m is mega 1 .6 megagrams and the center of mass g at g which is right here if the static question of friction between the shoulder of the tires and road is four determine the greater slope of the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at the constant velocity.
00:39
So let's um there's negative forces.
00:42
So we have a centrifugal force here, which is mb squared over r.
00:47
It's called fc for now.
00:49
Fc.
00:51
And sorry, the fc should be in this like this.
00:57
So to find the component which is in this direction, and you use the triangle completed, and this would be a heater.
01:07
So this side would be fc theta and this side usc sign theta now we have our weight acting here so the component of our weight which is here would be m g plus theta come out of our weight which is like this be mg sign theta so now our friction force will be acting here for the greatest amount of slope right because you want to extend this to the maximum value for the slopes so that the friction out force has to be the last line of defense in a sense so it has to oppose the motion down going down so now right on that force equation we can take right as a positive direction and left as the negative direction so we have our frictional force going to the right we have f plus theta, fc cost, theta, plus we have minus mg sine theta.
02:40
And this would be equal to zero because we don't want this to go up or down the slope.
02:47
So now for a frictional force, to resolve our forces in the y direction, can take up as positive and down as negative.
02:56
So our normal force is acting here.
02:59
So our normal force plus minus fc.
03:07
Sine theta minus m g plus theta is equal to zero so normal force we call to fc sine theta plus m g plus theta now so the frictional force is equal to mu r m m n medium normal force so if you're substitutes right here we have muir times f c sign let's just go ahead and substitute what f c is fc is the centrifugal force which is equal to mv squared of r see this mv squared of r sine theta plus m g plus plus theta plus mv squared of r plus theta minus m g sine theta is equal to zero and substitute question the friction is four our mass is 1 .6 megagrams which is 1 .6 times 10th to power of 6 grams now to convert grams to kilograms multiply by 10th of 3 minus 3 rather times the velocity we don't know the velocity let's live that as it is better by the radius so the problem with this is the question doesn't take into account the centrifugal force, which is just not practical...