A 6.99-kg block is placed on top of a 12.9-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.565. What is the maximum horizontal force that can be applied before the 6.99-kg block begins to slip relative to the 12.9-kg block, if the force is applied to (a) the more massive block and (b) the less massive block?

To find the maximum horizontal force, we need to consider the forces acting on the system. Let's start by analyzing the forces acting on the two blocks separately.

For the 12.9-kg block:
- The gravitational force acting downwards (weight) is given by F₁ = m₁ * g, where m₁ is the mass of the block and g is the acceleration due to gravity.
- The normal force exerted by the table upwards is equal to the weight of the block, which is F₁.

For the 6.99-kg block:
- The gravitational force is given by F₂ = m₂ * g, where m₂ is the mass of the block and g is the acceleration due to gravity.
- The force of static friction acting between the two blocks is μ * F₁, where μ is the coefficient of static friction and F₁ is the normal force exerted by the 12.9-kg block.

In this case, the maximum horizontal force that can be applied before the 6.99-kg block begins to slip relative to the 12.9-kg block is equal to the force of static friction.

(a) If the force is applied to the more massive block (12.9-kg block):
The maximum force of static friction is μ * F₁, so the maximum horizontal force that can be applied is 0.565 * F₁.

(b) If the force is applied to the less massive block (6.99-kg block):
The maximum force of static friction is μ * F₁, so the maximum horizontal force that can be applied is 0.565 * F₁.

To find the value of F₁, we need to equate the weight of the 12.9-kg block to the normal force exerted by the table. The weight is given by F₁ = m₁ * g, where m₁ is the mass of the 12.9-kg block and g is the acceleration due to gravity (approximately 9.8 m/s²). Thus, F₁ = (12.9 kg) * (9.8 m/s²).

By substituting the known values into the equations above, we can then calculate the maximum horizontal force that can be applied in each scenario.

To find the maximum horizontal force that can be applied before the 6.99-kg block begins to slip relative to the 12.9-kg block, we need to consider the forces acting on the blocks and the conditions for equilibrium.

(a) When the force is applied to the more massive block:
In this case, the two blocks will move together, and the maximum horizontal force will depend on the static friction between the blocks. The force of static friction can be calculated using the equation:

Fs = μs * N

Where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force between the blocks.

To calculate the normal force (N), we first need to determine the gravitational forces acting on each block. The force of gravity is given by:

Fg = m * g

Where m is the mass of the object and g is the acceleration due to gravity (~9.8 m/s^2).

For the 6.99-kg block:
Fg1 = m1 * g = 6.99 kg * 9.8 m/s^2

For the 12.9-kg block:
Fg2 = m2 * g = 12.9 kg * 9.8 m/s^2

The normal force (N) is equal to the force of gravity on the top block since there is no vertical acceleration:

N = Fg2

Now we can calculate the force of static friction (Fs):

Fs = μs * N
Fs = μs * Fg2

Substitute the values and calculate Fs.

(b) When the force is applied to the less massive block:
In this case, the normal force is not the same. To find the new normal force (N), we need to consider the force balance on the 6.99-kg block.

Force balance on the 6.99-kg block:
N - Fg1 = ma

Where N is the normal force, Fg1 is the force of gravity on the 6.99-kg block, m is the mass of the 6.99-kg block, and a is the acceleration of the system.

The force of static friction is still given by the equation:

Fs = μs * N

Substitute the value of N obtained from the force balance equation and calculate Fs.

So, (a) the maximum horizontal force is equal to Fs when applied to the more massive block, and (b) the maximum horizontal force is equal to Fs when applied to the less massive block.