A platform that rolls on wheels and has inertia mp is attached to a wall by a horizontal spring of spring constant k. A load of inertia ml sits on the platform, and the coefficient of static friction between platform and load is μs. If you pull the platform away from the wall so that the spring is stretched a distance x from its relaxed length and then let go, the platform-spring combination bounces back and forth.
If you want the load to stay in place held only by friction, what is the maximum distance xmax you can stretch the spring from its relaxed length and have the load stay on when you release the platform-spring combination?
To find the maximum distance xmax that you can stretch the spring and still have the load stay in place when you release the platform-spring combination, let's analyze the forces acting on the system.
When the platform-spring combination is released, it experiences a restoring force due to the spring. This force can be given by Hooke's Law:
F = -kx
Where F is the force, k is the spring constant, and x is the displacement from the relaxed length.
In order for the load to stay in place, the static friction between the platform and the load must provide an equal and opposite force to the restoring force of the spring. The maximum static friction force is given by:
F_max = μs * m * g
Where μs is the coefficient of static friction, m is the mass of the load, and g is the acceleration due to gravity.
Setting the maximum static friction force equal to the restoring force due to the spring:
μs * m * g = -k * x
Rearranging the equation to solve for x:
x = -(μs * m * g) / k
The negative sign indicates that the displacement x is in the opposite direction to the initial stretch.
Therefore, the maximum distance xmax you can stretch the spring from its relaxed length and still have the load stay in place when you release the platform-spring combination is given by the equation:
xmax = -(μs * m * g) / k
To find the maximum distance xmax you can stretch the spring from its relaxed length and have the load stay on when you release the platform-spring combination, you need to consider the forces acting on the load.
When the platform-spring combination is released, it will experience a restoring force due to the spring and frictional force due to the static friction between the platform and the load. The maximum distance xmax can be determined by balancing these forces.
1. Calculate the restoring force due to the spring:
The restoring force exerted by the spring is given by Hooke's Law: F_spring = -kx, where k is the spring constant, and x is the displacement of the spring from its relaxed length.
2. Calculate the maximum static frictional force:
The maximum static frictional force is given by F_friction_max = μs * m * g, where μs is the coefficient of static friction, m is the mass of the load, and g is the acceleration due to gravity.
3. Equate the restoring force and maximum static frictional force:
Set F_spring = F_friction_max, and solve for x:
-kx = μs * m * g
x = -(μs * m * g) / k
4. Find the absolute value of x to get the maximum distance xmax:
xmax = |-(μs * m * g) / k|
Note that the negative sign arises because when the spring is stretched, the displacement x is in the opposite direction of the restoring force.
So, the maximum distance xmax you can stretch the spring from its relaxed length and have the load stay on when you release the platform-spring combination is given by |-(μs * m * g) / k|.
max spring force = -k x
a = -k x/(mp+ml)
normal force between = ml g
max friction force = us ml g
that must result in acceleration a
a = us ml g/ml = -k x/(mp+ml)
so
us g =-k x/(mp+ml)