A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first block. The sur

Question

A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first block. The surfaces of the blocks are rough. A constant force of magnitude F is applied to the first block as shown in the figure. (a) Construct free-body diagrams for each block. (b) Identify the horizontal force that causes the block of mass m to accelerate. (c) Assume that the upper block does not slip on the lower block, and find the acceleration of each block in terms of m and F.

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Thiên Ân 1 month 2021-08-04T01:29:41+00:00 1 Answers 5 views 0

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    2021-08-04T01:31:08+00:00

    By Newton’s second law, assuming F is horizontal,

    • the net horizontal force on the larger block is

    Fµmg = 3mA

    where µmg is the magnitude of friction felt by the larger block due to rubbing with the smaller one, µ is the coefficient of static friction between the two blocks, and A is the block’s acceleration;

    • the net vertical force on the larger block is

    4mg – 3mgmg = 0

    where 4mg is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3mg is the weight of the larger block, and mg is the weight of the smaller block;

    • the net horizontal force on the smaller block is

    µmg = ma

    where µmg is again the friction between the two blocks, but notice that this points in the same direction as F. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

    • the net vertical force on the smaller block is

    mgmg = 0

    where mg is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

    (You should be able to draw your own FBD’s based on the forces mentioned above.)

    (c) Solve the equations above for A and a :

    A = (Fµmg) / (3m)

    a = µg

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