In a movie, a stuntman places himself on the front of a truck as the truck accelerates. The co efficient of friction between the stuntman and the truck is 0.65. The stuntman is not standing on anything but can “stick” to the front of the truck as long as the truck continues to accelerate. What minimum forward acceleration will keep the stuntman on the front of the truck?

Answer 1

FFirst, we analyse the Vertical axis, to keep the s-man from falling, the upward friction force must equal to the downward W, which is m*g. Thus, we derive the equation f=μ*Normal force=m*g=W

SSecond, let's focus on the horizontal axis, the force components on the horizontal axis are Normal force and applied force by the truck. but in this case, the Normal force IS the applied force. so the Fnet=Normal force=m*a.

now let's plug the "Normal force = m*a" from the SSecond part into the equation "μ*Normal force=m*g" from the FFirst part. so we get the equation "μ* m* a=m* g". both m cancel out, and we get "μ*a=9.8".
value of μ is given, which is 0.65. so finally "a= 9.8/0.65=15.0769...m/s^2"