help on physics problem!!?
A block of mass 1.74 kg is placed on top of a
second block of mass 6.51 kg. The coefficient
of kinetic friction between the 6.51 kg block
and the surface is 0.19. A horizontal force F
is applied to the 6.51 kg block.
Find the minimum coefficient of static friction
μt between the blocks such that the 1.74 kg
block does not slip under an acceleration of
4.21 m/s2.
Answers & Comments
Verified answer
First calculate the force required to accelerate the 1.74 kg block at 4.21 m/s^2.
F = ma
F = 1.74 kg * 4.21 m/s^2
F = 7.3 N
Now calculate the minimum μ to provide that much force:
Fμ = μN
where
Fμ = minimum friction force = F calculated above
μ = coefficient of friction
N = normal force
Fμ = μ*mg
7.3 N = μ*1.74 kg * 9.81 m/s^2
μ = 0.43
Here is an alternate method that skips a step and will show that the coefficient of static friction required is simply the proportion of lateral acceleration to g
F = ma
and
F = μmg
Set them equal:
μmg = ma
μ = a/g
μ = 4.21 / 9.81 = 0.43