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?
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"
To determine the minimum forward acceleration needed to keep the stuntman on the front of the truck, we can consider the forces acting on the stuntman.
1. The force of gravity: This force pulls the stuntman downward, but since he is not standing on anything, it does not have any effect on his motion in the horizontal direction.
2. The normal force: Since the stuntman is pressing against the front of the truck, the truck exerts a normal force on the stuntman in the upward direction. This normal force counteracts the force of gravity.
3. The force of friction: The coefficient of friction between the stuntman and the truck is given as 0.65. The force of friction is proportional to the normal force and opposes the motion of the stuntman.
Since the stuntman remains on the front of the truck as long as it continues to accelerate, we can assume that the force of friction is equal to the maximum possible force of static friction (before the stuntman starts to slip).
The equation for the maximum force of static friction (F_friction) is given by:
F_friction = coefficient of friction (µ) * normal force (N)
In this case, the normal force is equal to the force of gravity acting on the stuntman:
N = m * g
where m is the mass of the stuntman and g is the acceleration due to gravity.
Now, the force of friction can also be expressed as:
F_friction = m * a
where a is the acceleration of the stuntman.
By equating these two expressions for the force of friction, we can find the minimum acceleration required to keep the stuntman on the front of the truck:
m * a = µ * m * g
Simplifying the equation:
a = µ * g
Substituting the given values:
a = 0.65 * 9.8 m/s^2
Therefore, the minimum forward acceleration required to keep the stuntman on the front of the truck is approximately 6.37 m/s^2.