the
1f=20N down
R=86.6N right
A force of 100 N is applied on a stationary block of mass 3kg as shown in figure. I f the coefficient of frictioon between the surface and the block is 0.25
what is the magnitude anddirection of frictional force
what is the magnitude and direction of the normal reaction
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UP 0 DOWN 0 0 6
6 Answers
62mg = 3.10 = 30N
F. sin(30) = 100/2 =50
clearly we know that if friction was not acting the acceleration wud have been upwards. Friction will tend to oppose that motion...
So friction will be downwards
Now we take teh max friction force possible = 0.25 F cos (30) = 25.√3/2 = 21.65 > 20 (which is the unbalanced force)
Hence the frictional force acting will be 20N
62Normal reaction is perpendicular to the surface of contact!
Its magnitude is F cos (30) = 100. cos 30
The posiiton of this might not pass through the center of mass! (But I dont think that is being asked here.. bcos in that case the dimension of the block wud have been given!)