Ksuweb.kennesaw.edu
Problem 4First write down everything given, everything assumed, and what we’re looking for.Given: v= 25.0 m/sradius =125 mμ= 0.558mass is not given but we’ll see that it’s unimportantAssumptions: acceleration due to gravityFind: Whether you will fall off and max velocity that you won’t fall off.In order to save time we will just focus on the second answer. Because if the max velocity is greater than 25.0 m/s then we will stay on the truck at 25.0 m/s. If max velocity is less than 25.0 m/s the opposite is true.Let’s make some pictures. First one from above.Vmax125 mNow let’s make one from the front of the truck. We need to take all our forces into consideration.mgFrictional forceCentripetal forceNormal forceSo in order for you to stay on the truck your frictional force must be greater than or equal to your centripetal force or (Frictional force) ≥ (Centripetal force). We define frictional force as our normal force times our coefficient of friction. Because we have no acceleration in the y direction our y forces cancel out and normal force is equal to mg. Our centripetal force is defined as m(v2/r). We can say that you will stay on the truck when (μmg)≥(m(v2/r)). Our m’s cancel so the equation is now (μg)≥(v2/r). If we rearrange terms we get v≤√μgr. So v≤√(0.558)(9.81 m/s2)(125 m) or v≤ 26.6 m/s.The highest value our v can be while still making the above true is when v is 26.6 m/s. Because 25.0 m/s is less than our max v we would stay on the truck. ................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.