Don't try this with your parents' dishes

By definition, the coefficient of static friction $\mu_s$ is given by the formula $\mu_s = \frac{F}{N}$ , where $F$ is the force required to set it in motion. $F=ma$ and $N$ is the normal force $N=mg$ . Thus, by substitution, $\mu_s = \frac{F}{N} = \frac{ma}{mg} = \frac{a}{g}$ . By rearrangement, $a = g\mu_s = 9.8\left(0.3\right) = 2.94$ .

Suppose the mass of all stack on the cloth is 'M'. The stack will exert a force Mg downward; this will be equal to the normal reaction by the table (N).

The co-efficient of friction is μ.

If the force exerted is towards right, Force of friction is towards left. Force of friction exerted will be equal to μN.

By Newton's second law, F=Ma μN=Ma μMg=Ma

=> a= μg

μ is given as 0.3, and g=9.8 m/s^2

=> a=0.3*9.8= 2.94 m/s^2

Consider the situation form the point of view of the tablecloth. We can say that we want to give the plates an acceleration so they move, not the cloth. This then becomes trivial to solve.

Let the mass of the plates be $m$ , then the force of friction is $mg\mu$ . A little more force than this will be enough to accelerate the plates. We can then divide this by $m$ to get the acceleration which equals $g\mu = 9.8\times 0.3 = \boxed{2.94}$

$a(f) = u \times a$

Where:

$a(f)$ = needed acceleration applied to the table cloth so the plates start to slip.

$u$ = coefficient of friction

$a$ = acceleration due to gravity

$a(f) = 0.3 \times (9.8m/s^2)$ $= 2.94m/s^2$

Because the static friction is given, all you need to do is to multiply it to the acceleration due to gravity. This will give you the acceleration needed in this given scenario. 0.3 X 9.8 = 2.94

The acceleration of the tablecloth will be given (a)=u g=0.3 9.8=2.94ms^-2.

Let F = frictional force; c = coefficient of static friction; N = normal force; m = mass; g = acceleration due to gravity = 9.8 m/s^2 and W = weight.

Therefore,

F = cN and W = N = mg

F = cmg

ma=cmg

a=cg

Thus, a = 0.3*9.8 = 2.94 m/s^2 :)

The downwards force is the weight of the plates (w) which is mass of the plates times the acceleration due to gravity $w=-mg$ This is the only downward force acting on the plates so the normal force (N) is equal and opposite to w so $N=mg$ We also know that the frictional force (f) where mu is the coefficient of friction can vary up to a certain limit $f\leq \mu N$ $f\leq \mu m g$ The horizontal force (F) of the pull will be given by $F=ma$ When the plates just start to move this will be where the frictional force is at its maximum i.e. $f=\mu mg$ These two forces f and F will be equal when the plates is begin to move so $F=f$ $ma=\mu mg$ $a=\mu g$ $a=0.3 \cdot 9.8ms^{-1}$ $a=2.94ms^{-1}$

Consider the situation form the point of view of the tablecloth. We can say that we want to give the plates an acceleration so they move, not the cloth. This then becomes trivial to solve.

Let the mass of the plates be m, then the force of friction is mgμ. A little more force than this will be enough to accelerate the plates. We can then divide this by m to get the acceleration which equals gμ=9.8×0.3=2.94

After drawing the FBD, we know that N (normal force) = mg.

Ff= u(N)

=u(mg)

Force Applied>Ff

Fa > 0.3mg

ma > 0.3mg

a>0.3(9.8)

a>2.94

Seeming as the weight of the plate will be $W=mg$ directly down, the normal force to the plate will be equal to the weight directly up. The force of friction can be represented as $F=kN$ where $k$ is the coefficient of friction and $N$ is the normal force; hence $F=0.3 \times 9.8 \times m$ horizontally. Futhermore, the required force to start moving the plate will be infinitesimally larger than the force of friction, therefore the force required to make the plate slip is $F=2.94m$ . Seeming as force is equal to $F=ma$ , this implies that the acceleration required has a magnitude of $2.94$ .