The coefficient of static friction between block \(\mathrm{A}\) and a horizontal floor is 0.45 and the coefficient of static friction between block \(\mathrm{B}\) and the floor is \(0.30 .\) The mass of each block is $2.0 \mathrm{kg}$ and they are connected together by a cord. (a) If a horizontal force \(\overrightarrow{\mathbf{F}}\) pulling on block \(\mathrm{B}\) is slowly increased, in a direction parallel to the connecting cord, until it is barely enough to make the two blocks start moving, what is the magnitude of \(\overrightarrow{\mathbf{F}}\) at the instant that they start to slide? (b) What is the tension in the cord connecting blocks \(A\) and \(B\) at that same instant?

Before we dive into calculations, let us first understand the forces acting on the blocks. The most important ones are gravitational force, normal force, force due to the tension in the cord, and the horizontal pulling force. To find the force needed to make the blocks start moving, we will first find the maximum static friction forces between blocks A and B and the floor.

The maximum static friction force acting on an object can be calculated using the formula: $$f_{s(max)} = μ_s × N$$ where \(f_{s(max)}\) is the maximum static friction force, \(μ_s\) is the coefficient of static friction, and \(N\) is the normal force. The normal force for both blocks is equal to their respective gravitational forces since they are on a horizontal surface: $$N_A = m_A × g$$ $$N_B = m_B × g$$ where \(m_A\) and \(m_B\) are the masses of blocks A and B, and \(g\) is the acceleration due to gravity. Substituting the given values, we get: $$N_A = N_B = (2.0\,\text{kg}) × (9.81\,\text{m/s}^2) = 19.62\,\text{N}$$ Now, we can calculate the maximum static friction forces for both blocks: $$f_{s(max)A} = μ_{sA} × N_A = (0.45) × (19.62\,\text{N}) = 8.829\,\text{N}$$ $$f_{s(max)B} = μ_{sB} × N_B = (0.30) × (19.62\,\text{N}) = 5.886\,\text{N}$$

Let \(F_A\) and \(F_B\) be the frictional forces acting on block A and block B, respectively. When the pulling force (\(\overrightarrow{\mathbf{F}}\)) is just enough to make the blocks start moving, both frictional forces will be at their maximum magnitudes. So, we can write: $$F_A = f_{s(max)A}$$ $$F_B = f_{s(max)B}$$ As block B is connected to block A, the force that makes block A move must equal the pulling force minus the friction force on block B: $$F_A = F - F_B$$ Plugging in the maximum static friction forces, we get: $$F_A = F - f_{s(max)B}$$ $$8.829\,\text{N} = F - 5.886\,\text{N}$$ Now, we can solve for the magnitude of the pulling force: $$F = 8.829\,\text{N} + 5.886\,\text{N} = 14.715\,\text{N}$$ So, the magnitude of the pulling force needed to make the blocks start moving is \(14.715\,\text{N}\).

At the instant the blocks start sliding, the system reaches an equilibrium. As a result, the tension in the cord connecting the blocks must be equal to the friction force acting on block A: $$T = F_A$$ Since \(F_A = f_{s(max)A}\), we have: $$T = 8.829\,\text{N}$$ Thus, the tension in the cord connecting blocks A and B at the instant they start sliding is \(8.829\,\text{N}\).