Block A of mass 2.2 kg is on block B of mass 5.2 kg as shown in the above figure. The lower block is on a frictionless surface while the coefficient of static friction is 0.3 between the two blocks.

a) If they are moving at constant velocity, what is the frictional force between A and B?
b) What is the maximum horizontal force that can be applied to B without A slipping?

I tried for a) Ff= 0.3*2.2*g = 6.4746N (both positive and negative value) but it was incorrect, and b) I tried 0.3g(2*2.2+5.2) = 28.224 was also incorrect.

To solve this problem, we need to first understand the concept of friction and how it relates to forces and motion.

a) If the blocks are moving at a constant velocity, it means that the net force on the system is zero. In this case, the frictional force between block A and block B must be equal in magnitude and opposite in direction to the external force applied to the system.

To find the frictional force, we can use the equation:

Ff = μs * N

where Ff is the frictional force, μs is the coefficient of static friction, and N is the normal force.

The normal force is equal to the weight of block A, which can be calculated as:

N = m * g

where m is the mass of block A and g is the acceleration due to gravity.

Using the values provided:
m = 2.2 kg and g = 9.8 m/s^2, we can calculate the normal force:

N = 2.2 kg * 9.8 m/s^2 = 21.56 N

Substituting this value into the equation for frictional force:

Ff = 0.3 * 21.56 N = 6.468 N

So, the frictional force between block A and block B is approximately 6.468 N.

b) The maximum horizontal force that can be applied to block B without block A slipping can be found by considering the conditions for static equilibrium. When the blocks are at rest, the maximum force that can be applied without causing slip is equal in magnitude to the maximum static friction.

The maximum static friction can be calculated using the equation:

Ff(max) = μs * N

where Ff(max) is the maximum static friction, μs is the coefficient of static friction, and N is the normal force, which is the weight of block A.

Using the same values provided earlier:

N = 2.2 kg * 9.8 m/s^2 = 21.56 N

Substituting into the equation for maximum static friction:

Ff(max) = 0.3 * 21.56 N = 6.468 N

Therefore, the maximum horizontal force that can be applied to block B without block A slipping is approximately 6.468 N.

Let's solve each part step-by-step:

a) To find the frictional force between blocks A and B when they are moving at a constant velocity, we need to consider that the frictional force opposes their relative motion.

We know that the coefficient of static friction between the two blocks is 0.3. However, when they are moving at a constant velocity, they have overcome the static friction, and we instead need to consider the coefficient of kinetic (dynamic) friction.

The frictional force (F_f) can be calculated using the equation:

F_f = μ_k * N

where μ_k is the coefficient of kinetic friction, and N is the normal force between the two blocks.

In this case, since the lower block is on a frictionless surface, the normal force is simply the weight of block A, which is equal to its mass (m_A) multiplied by the acceleration due to gravity (g).

Therefore, the frictional force (F_f) can be calculated as:

F_f = μ_k * m_A * g

Plugging in the given values, we have:

F_f = 0.3 * 2.2 kg * 9.8 m/s^2

Calculating this, we find:

F_f ≈ 6.47 N

So the correct value for the frictional force between blocks A and B, when they are moving at a constant velocity, is approximately 6.47 N.

b) To find the maximum horizontal force (F_max) that can be applied to block B without block A slipping, we need to consider the maximum static friction that can be applied.

The maximum static friction (F_s) can be calculated using the equation:

F_s = μ_s * N

where μ_s is the coefficient of static friction and N is the normal force between the two blocks.

In this case, the normal force (N) is equal to the weight of block A plus the weight of block B, which is equal to the sum of their masses (m_A and m_B) multiplied by the acceleration due to gravity (g).

Therefore, the maximum static friction (F_s) can be calculated as:

F_s = μ_s * (m_A + m_B) * g

Plugging in the given values, we have:

F_s = 0.3 * (2.2 kg + 5.2 kg) * 9.8 m/s^2

Calculating this, we find:

F_s ≈ 17.16 N

So the correct value for the maximum horizontal force that can be applied to block B without block A slipping is approximately 17.16 N.

I hope this helps! Let me know if you have any further questions.