The cake company truck is on it's way deliver a cake for a party when it rounds a curve of radius 20 m at a speed of 12 m/s. What coefficient of friction is needed between the cake pan and the truck in order to keep the pan from slipping?

please show work!

To avoid slipping, the maximum static friction force,

M g Us,
must exceed the centripetal force
M V^2/R

Set them equal and solve for the static friction coefficient, Us.

Us = V^2/(R*g)

That will be the minimum value required.

To determine the coefficient of friction needed between the cake pan and the truck to keep the pan from slipping, we can use the concept of centripetal force.

The centripetal force required to keep the cake pan from slipping can be calculated using the equation:

F = m * (v^2 / r)

Where:
F is the centripetal force,
m is the mass of the cake pan,
v is the velocity of the truck, and
r is the radius of the curve.

In this case, the cake pan is not slipping, which means that the frictional force between the pan and the truck provides the necessary centripetal force. The frictional force can be represented as:

F_friction = μ * N

Where:
μ is the coefficient of friction, and
N is the normal force between the pan and the truck.

Since the truck is not sliding in this situation, the normal force is equal to the weight of the pan, which can be calculated as:

N = m * g

Where:
g is the acceleration due to gravity (approximated as 9.8 m/s^2).

We can substitute N in the equation for F_friction to get:

F_friction = μ * m * g

Since the frictional force provides the centripetal force, we can equate the two equations:

F = F_friction

m * (v^2 / r) = μ * m * g

Simplifying the equation by canceling out the mass (m) and solving for μ, we find:

μ = (v^2 / (r * g))

Given that the radius of the curve is 20 m, the speed of the truck is 12 m/s, and the acceleration due to gravity is 9.8 m/s^2, we can substitute these values into the equation to find the coefficient of friction (μ):

μ = (12^2 / (20 * 9.8))
= 1.469

Therefore, the coefficient of friction needed between the cake pan and the truck to keep the pan from slipping is approximately 1.469.