Question

For the system below where m1=45kg and the coefficient of statice friction between m1 and the table is .40, how large must m2 be made without the system starting into motion? Not sure how to start, help would be awesome, thanks!

Answer 1

They're asking what mass M2 should be so that it's as large as possible but the M2 block does not fall (which means the M1 block doesn't move either)

Static friction is holding M1 in place. The largest amount of static friction to overcome is equal to μ∗η


greek letter mu = μ = coefficient of static friction = 0.4
greek letter eta = η = normal force = m*g = 45*9.8 = 441

the normal force is approximate

So the maximum force of static friction is equal to μ∗η=0.4∗441=176.4

This is the force to overcome. If the force pulling down on M2 is larger than 176.4, the whole system will move. If the force pulling down on M2 is 176.4 roughly, then we're at the max limit. It's the largest amount of force allowed where the system is stationary

So we have the force (176.4) and we know g = 9.8 approx

so,

F = m*g
176.4 = m*9.8
176.4/9.8 = m*9.8/9.8
18 = m
m = 18

So it looks like if M2 is 18 kg, then the system stays still and this is the largest M2 can get. Any larger and the system will move.