Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box of mass M. There is friction between the surfaces of blocks 2M and 3M (μs = 0.26, μk = 0.15) so the three blocks accelerated together to the right.

What is the acceleration of the blocks? (You may assume block 3M does not slide or fall off block 2M.)

What is the maximum force F that can be applied, before the 3M block slides off?

In your question where as ask to calculate the acceleration of the block base on the figure that is shown in you diagram. Base on my calculation and through step by step procedure and with the guidance of the theories and by the use of formulas, I came up with an answer of F>3μtMg

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aristocles aristocles · Answer 2 · 2019-10-23T20:45:40-04:00

Answer:

Part a)

\(a = 0.67 m/s^2\)

Part b)

\(F_{max} = 107.1 N\)

Explanation:

Part a)

Applied force on the block system is given as

\(F = 28 N\)

total mass of the system is given as

\(m = M + 3M + 2M\)

\(m = 6M\)

now acceleration of the system is given as

\(a = \frac{F}{m}\)

\(a = \frac{28}{6M}\)

\(a = 0.67 m/s^2\)

Part b)

When 3M mass block just start sliding then the force of friction on it will reach to its maximum value

\(F = F_s\)

\(F = \mu_s (3M)(g)\)

\(F = 0.26(3\times 7)(9.81)\)

now the acceleration of the block is given as

\(a = \frac{F}{3M} = \mu_s g\)

\(a = 0.26(9.81)\)

\(a = 2.55 m/s^2\)

now we have

\(F_{max} = (M + 2M + 3M) a\)

\(F_{max} = 6(7)(2.55)\)

\(F_{max} = 107.1 N\)