A stack of papers in the rear of a delivery truck moving at 20m/s has a 20kg mass. What is the minimum radius the truck can turn for the papers not to slip if coefficient of static friction is 0.3?
To find the minimum radius that the truck can turn without the papers slipping, we need to consider the forces acting on the papers.
First, let's determine the maximum static friction force that can be exerted on the papers before they start to slip.
The formula for static friction force is given by:
Fs = μs * N
where Fs is the static friction force, μs is the coefficient of static friction, and N is the normal force.
In this case, the normal force acting on the papers is equal to the weight of the papers, which is given by:
N = m * g
where m is the mass of the papers and g is the acceleration due to gravity (approximately 9.8 m/s^2).
N = 20 kg * 9.8 m/s^2 = 196 N
Now, we can calculate the maximum static friction force:
Fs = 0.3 * 196 N = 58.8 N
The maximum static friction force provides the necessary centripetal force for the papers to stay in place while the truck is turning. The centripetal force is given by:
Fc = m * v^2 / r
where Fc is the centripetal force, m is the mass of the papers, v is the velocity of the truck, and r is the radius of the turn.
We can rearrange this equation to solve for the radius:
r = m * v^2 / Fc
Since the maximum static friction force (58.8 N) is providing the centripetal force, we can substitute that into the equation:
r = (20 kg * (20 m/s)^2) / 58.8 N
Calculating this gives:
r = 136.05 m
Therefore, the minimum radius the truck can turn for the papers not to slip is approximately 136.05 meters.