1. The problem statement, all variables and given/known data A 3500kg truck is carrying an unsecured 400kg box containing glass. The coefficient of static friction between the box of glass and the truck is 0.50. The coefficient of rolling friction between the truck's tires and the road is 0.03. The coefficient of static friction between the truck's tires and the road is 0.95. The truck comes to an unbanked curve in the road with a 40m radius. What is the maximum speed the truck can take the curve without damaging the glass, assuming the glass is damaged should the box slide on the truck, or the truck itself should slide during the curve. 2. Relevant equations I assume: a=(v^2)/r F(friction)=(coefficient)*F(normal) 3. The attempt at a solution I'm not completely familiar with the concept, but as I understand it, the car assumes strictly static friction (for my purpose) while attempting the curve, thus the coefficient of rolling friction is completely null, I think. What I'm not certain with is how exactly the box will react with the truck, being unattached. The only lateral force on the truck attempting the curve is it's static friction which would be: F(static friction of truck to road)=0.95*((3500kg+400kg)*9.8m/s^2)=36309N Surely the box's mass would combine with the truck's? But in regards to its being damaged by its own sliding, I assume the same static friction applied to the car is also affecting the box: F(static friction of box to truck)=0.50*(400kg*9.8m/s^2)=1960N If it holds true the box will undergo the same force, then you could just ignore the whole system of the car, easily recognizing that maximizing it's friction would completely compromise the box's limits of friction, thus just calculating the box's limits: 1960N=(400kg*v^2)/40m v=14m/s before the box begins to slide, whereas v=19.29m/s before the car begins to roll. Thus 14m/s is the maximum speed. Sorry if it sounds confusing. It doesn't sound very sound to me, either. I guess I'm searching for a confirmation of my logic.