We do not know yet if the top block slips.
First, assume it does not slip.
F = m a
30 = (4+2) a
a = 5 m/s^2
Now
How much force is required to accelerate the top block at 5 m/s^2?
F = 2 (5) = 10 N required
How much do we have?
Fnormal = m g = 2*9.81 = 19.62 N
so friction max = .8 * 19.62 = 15.7 N
But we only need 10 N to keep that top block stuck on the bottom one, so it does not slide and we are finished.
4 kg block rests on a smooth horizontal surface,upper surface of 4 kg is rough.A block of mass 2 kg is placed on its upper surface.The acceleration of upper block with respect to earth when 4 kg mass is pulled by a horizontal force of 30N?coefficient of static friction=0.8 and coefficient of kinetic friction=0.6 I have been working to solve this but not getting it as i do not know what is force of friction on 2 kg block .I know it is coefficient of friction divided by normal reaction but what would be normal reaction?
1 answer
1 answer