please explain in details the steps that led to the anwer below. how did ug(m1+m

Question

please explain in details the steps that led to the anwer below. how did ug(m1+m1+m2) . please help.

A 4.00-kg block rests between the floor and a 3.00-kg block as shown in the figure below. The 3.00-kg block is tied to a wall by a horizontal rope. If the coefficient of static friction is 0.800 between each pair of surfaces in contact, what force must be applied horizontally to the 4.00-kg block to make it move? 23.5 N 16.2 N 21.1N 54.9N 78.5N Solution F - F_fr_1 - F_fr_2 = ma_x F - mu F_N_1 - mu F_N_2 = 0 F = mu g(m_1 + m_1 + m_2) F = (0.8) (9.8 m/s^2) (2(3.0 kg) + 4.0 kg) = 78.4 N

Explanation / Answer

Here the upper block can't move because it is connected to string.

So the friction force on the upper block will be towards the right.

So on the lower block friction due to the upper block will be towards the left.

Similarly the frcition on the lower block due to the surface will be towards left.

So equation of force balance for the lower block will be:

F - f1r - f2r = 4*a

Now suppose the block is not slipping and the friction force has its maximum value.

The maximum value of static friction is normal times the friction coefficient.

Normal between block = 3*9.81 N

Normal between the surface and lower block = (4+3)*9.81 N

Hence F - 0.8*3*9.81-0.8*(4+3)*9.81 = 0

F = 78.48 N

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