A crate of mass 97 kg is loaded onto the back of a flatbed truck. The coefficient of static friction between the box and the truck bed is 0.26. What is the smallest radius of curvature (in m) that the truck can take, if the speed with which it is going around a circle is 44 m/s?
I'm not quite sure on how to solve this :S
To solve this problem, we can analyze the forces acting on the crate as the truck goes around a curve. The key forces involved are the static friction force and the centripetal force.
The maximum static friction force Fs(max) that can act on the crate can be found using the coefficient of static friction (μs) and the normal force (N) acting on the crate. The normal force is equal to the weight of the crate, which in this case is mass (m) times the acceleration due to gravity (g):
N = m * g
The maximum static friction force can then be calculated as:
Fs(max) = μs * N
To keep the crate from sliding on the truck bed, the static friction force must be equal to or greater than the centripetal force. The centripetal force (Fc) required to keep an object moving in a circular path can be calculated using the following equation:
Fc = m * v^2 / r
Where:
m = mass of the crate
v = velocity of the truck
r = radius of curvature of the truck's path
Setting the maximum static friction force equal to the centripetal force, we can solve for the radius of curvature (r):
μs * N = m * v^2 / r
Plugging in the given values:
μs = 0.26
m = 97 kg
v = 44 m/s
g = 9.8 m/s^2
N = m * g = 97 kg * 9.8 m/s^2
Now we can solve for r:
0.26 * (97 kg * 9.8 m/s^2) = 97 kg * (44 m/s)^2 / r
Simplifying the equation:
r = (97 kg * (44 m/s)^2) / (0.26 * (97 kg * 9.8 m/s^2))
After evaluating the above expression, we find that the smallest radius of curvature the truck can take is approximately 163.98 meters (or 164 meters when rounded to two decimal places).