SOLUTION: Determine the magnitude and direction of acceleration of a system composed of a 4kg cast iron block and a 3kg steel block connected by a pulley. The two blocks are placed on each s

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Question 1182653: Determine the magnitude and direction of acceleration of a system composed of a 4kg cast iron block and a 3kg steel block connected by a pulley. The two blocks are placed on each side of a 2-sided brass inclined plane. The cast iron is placed on the 20-degree side, while the steel is placed on the 45-degree side of the inclined. The coefficients of kinetic friction for cast iron and steel on brass are 0.30 and 0.44 respectively. The static coefficients of friction are 0.40 for cast iron on brass and 0.50 for steel on brass.
Answer by ikleyn(52846) About Me  (Show Source):
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Determine the magnitude and direction of acceleration of a system composed of a 4kg cast iron block
and a 3kg steel block connected by a pulley. The two blocks are placed on each side of a 2-sided brass inclined plane.
The cast iron is placed on the 20-degree side, while the steel is placed on the 45-degree side of the inclined.
The coefficients of kinetic friction for cast iron and steel on brass are 0.30 and 0.44 respectively.
The static coefficients of friction are 0.40 for cast iron on brass and 0.50 for steel on brass.
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Let's compute some acting forces to determine what will happen with the system.


The rolling force acting to the 4 kg cast iron block is  F%5B1%2Cr_iron%5D = mg*sin(20°) = 4*10*0.342 = 13.68 newtons.

The static friction force for this block is  F%5B1%2Cstat_fr%5D = k%5Bstat_fr%5D%2Amg%2Acos%2820%5E0%29 = 0.4*4*10*0.940 = 15.04 newtons.



The rolling force acting to the 3 kg steel block is  F%5B2%2Cr_steel%5D = mg*sin(45°) = 3*10*0.707 = 21.21 newtons.

The static friction force for this block is  F%5B2%2Cstat_fr%5D = k%5Bstat_fr%5D%2Amg%2Acos%2845%5E0%29 = 0.5*4*10*0.707 = 14.14 newtons.



Comparing these forces, we see that the difference of acting rolling forces (21.21-13.68) = 7.53 newtons is NOT ENOUGH

to overcome the total static friction force on both blocks  (15.04 + 14.14) = 29.18 newtons.



So, the conclusion is that the system will REMAIN to be unmovable.

Solved.