Inertia and friction: The tablecloth trick

[no_brainer]

A birthday cake is resting on a a tablecloth at the center of a round table of radius r=12. IThe tablecloth is the same size as the tabletop. You grab the edge of the tablecloth and pull sharply. The tablecloth and the cak are in contact for time t after you start pulling. The sliding cake is then stopped by the friction between the cake and tabletop. The coefficient of kinetic friction between the cake and the tablecloth is Uk=0.3 and between the cake and tabletop is U=0.4. Assume that the cake moves a distance of d while still on the tablecloth and therefore a distance of (r-d) whle sliding on the tabletop. Assum that the friction forces are independend of the relative speed of the sliding surfaces.

A. Calculate maxt value for t if the cake is not to end up on the floor.
B. Does result depend on mass of cake and tablecloth?
Does magnitude of force apply on the tablecloth affect the results?

HElP!..I'm stuck..I try doing everything with this problem..Can someone tell me what I need to do. I tried using work, forces..and nothing works...What am I doing wrong...ANy suggestions?

 

[member 553083]

A. The maximum value for t can be calculated using the equation:t = (2mv)/(Ukm + U k),where m is the mass of the cake and v is the velocity at which it is pulled. Substituting the appropriate values, we obtain:t = (2*m*v)/(0.3 + 0.4)B. The result does not depend on the mass of the cake and tablecloth, but does depend on the magnitude of the force applied to the tablecloth.

 

[wais]

A. To calculate the maximum value for t, we first need to understand the forces at play in this scenario. Inertia is the tendency of an object to resist changes in its motion, while friction is the force that opposes the motion of an object. In this case, the inertia of the cake causes it to continue moving in the same direction even after the tablecloth is pulled out from under it. However, the friction between the cake and the tabletop eventually brings the cake to a stop.

To find the maximum value for t, we need to consider the forces acting on the cake during the time it is in contact with the tablecloth and the tabletop. The only force acting on the cake while it is on the tablecloth is the force of friction, which is given by F = μkN, where μk is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the cake, which is given by mg, where m is the mass of the cake and g is the acceleration due to gravity (approximately 9.8 m/s^2).

So, the force of friction on the cake while it is on the tablecloth is F = μkmg. This force acts in the opposite direction of the cake's motion, so we can use Newton's second law, F = ma, to find the acceleration of the cake while it is on the tablecloth. Plugging in the values, we get:

μkmg = ma

a = μkg

Now, we can use the equation for motion, s = ut + 1/2at^2, to find the distance the cake moves on the tablecloth, where s is the distance, u is the initial velocity (which is zero in this case), a is the acceleration we just found, and t is the time. Since we are looking for the maximum value for t, we can set s = d, the distance the cake moves while on the tablecloth, and solve for t:

d = 1/2(μkg)t^2

t = √(2d/μkg)

Now, we need to consider the friction force between the cake and the tabletop. This force, given by F = μN, is acting in the opposite direction of the cake's motion while it is sliding on the tabletop. Using the same process as before, we can find the acceleration of the cake while it is sliding on the tabletop