A crate of mass 132 kg is loaded onto the back of a flatbed truck. The coefficient of static friction between the box and the truck bed is 0.22. What is the smallest radius of curvature that the truck can take without the crate slipping, if the speed with which it is going around a circle is 44 m/s?
10
To determine the smallest radius of curvature that the truck can take without the crate slipping, we need to consider the forces acting on the crate.
The force of static friction between the crate and the truck bed provides the centripetal force necessary to keep the crate moving in a circle. The formula for centripetal force is:
F_c = m * a_c
Where F_c is the centripetal force, m is the mass of the crate, and a_c is the acceleration towards the center of the circle.
The acceleration towards the center of the circle can be calculated using the equation:
a_c = v^2 / r
Where v is the velocity of the truck and r is the radius of curvature of the circular path.
Since the crate is at the verge of slipping, the maximum force of static friction can be calculated using the formula:
F_static_max = μ * N
Where μ is the coefficient of static friction and N is the normal force. Since the crate is on a flatbed truck, the normal force is equal to the weight of the crate.
Now, we can combine these equations to find the smallest radius of curvature:
F_c = m * (v^2 / r) [equation 1]
F_static_max = μ * N [equation 2]
Since F_c and F_static_max are the same force, we can equate them:
m * (v^2 / r) = μ * N
Substituting N with m * g (where g is the acceleration due to gravity):
m * (v^2 / r) = μ * m * g
Canceling out the mass from both sides:
v^2 / r = μ * g
Rearranging the equation to solve for the radius:
r = v^2 / (μ * g)
Plugging in the given values:
v = 44 m/s
μ = 0.22
g = 9.8 m/s^2
r = (44^2) / (0.22 * 9.8)
r = 8744 / 2.156
r ≈ 4050.46 m
Therefore, the smallest radius of curvature that the truck can take without the crate slipping is approximately 4050.46 meters.