Can a Teacup Stay on a Rapidly Pulled Tablecloth?

[BrandonW]

A student pulls a tablecloth out from under a teacup. Suppose the tablecloth has mass m = 0.1 kg, the teacup has mass M = 0.2 kg, and the coefficient of kinetic friction between the tablecloth and the teacup, and between the tablecloth and the table, is μ = 0.4. The cup starts out 1 m from the edge of the tablecloth, and the student pulls hard enough to give the tablecloth a constant acceleration of 20 m/s2. Draw a free-body diagram for the cup, and another for the tablecloth. Calculate the distance the cup moves (relative to the table) before it reaches the edge of the tablecloth, the distance the tablecloth moves, and the speed of the cup just as it reaches the edge of the tablecloth.

The trick is in the constant acceleration I think.?

 

[jbsteele]

Free Body Diagram for Cup:The forces acting on the cup are the gravitational force, the normal force of the tablecloth, and the frictional force between the tablecloth and the cup. Fnet = Fg + Fn + Ffm*a = Fg + Fn + Ff Free Body Diagram for Tablecloth:The forces acting on the tablecloth are the gravitational force, the normal force of the table, and the frictional force between the tablecloth and the table. Fnet = Fg + Fn + Ffm*a = Fg + Fn + Ff Distance the cup moves (relative to the table) before it reaches the edge of the tablecloth:Using the equation of motion, v2 = u2 + 2as, we can calculate the distance the cup moved:d = (v2 - u2)/2a = [(20 m/s)2 - (0 m/s)2]/2(20 m/s2) = 100 m Distance the tablecloth moves:Using the equation of motion, x = ut + 1/2at2, we can calculate the distance the tablecloth moved:d = ut + 1/2at2 = 0 m + 1/2(20 m/s2)(100 m)2 = 100,000 m Speed of the cup just as it reaches the edge of the tablecloth:Using the equation of motion, v2 = u2 + 2as, we can calculate the speed of the cup:v2 = u2 + 2as = (0 m/s)2 + 2(20 m/s2)(100 m) = 4000 m2/s2 = 40 m/s

 

[baba89]

I would like to clarify that this question is not a mechanics question, but rather a physics question. Mechanics is a branch of physics that deals with the study of motion and forces.

To answer this question, we first need to draw free-body diagrams for both the teacup and the tablecloth. The free-body diagram for the teacup would show the force of gravity acting downwards (mg), the normal force from the table acting upwards (N), and the frictional force (F) acting in the opposite direction of motion. The free-body diagram for the tablecloth would show the force of gravity acting downwards (Mg), the normal force from the table acting upwards (N), and the frictional force (F) acting in the opposite direction of motion.

Using Newton's second law, we can write the following equations for both the teacup and the tablecloth:

Teacup: ΣF = ma
Tablecloth: ΣF = Ma

Since we know that the tablecloth has a constant acceleration of 20 m/s^2, we can substitute this value in the above equation for the tablecloth:

Tablecloth: ΣF = M(20)

We also know that the frictional force is equal to the coefficient of kinetic friction (μ) multiplied by the normal force (N):

F = μN

Substituting this into the equation for the teacup, we get:

Teacup: mg - μN = ma

We can also substitute the value of the normal force (N) from the equation for the tablecloth into the equation for the teacup:

Teacup: mg - μ(Mg) = ma

Solving for N, we get:

N = mg/(1+μ)

Substituting the value of N into the equation for the teacup, we get:

Teacup: mg - μ(mg/(1+μ)) = ma

Simplifying this equation, we get:

Teacup: a = g(1-μ)/(1+μ)

Using the given values for the mass of the teacup (M = 0.2 kg), the mass of the tablecloth (m = 0.1 kg), and the coefficient of kinetic friction (μ = 0.4), we can calculate the acceleration of the teacup:

a = 9.8 m/s^2 (1-0.4