friction problem?

Question

Consider two boxes being pulled by a cord across a floor with friction. The lower box has mass m₂ and a co-efficient of kinetic friction μ(k) with the floor. The upper box has mass m₁ and a co-efficient of static friction μ(s) with the lower box. The cord makes an angle Ѳ with the horizontal. Assume that the angle is small enough that m₂ remains in contact with the floor.



1) What is the maximum acceleration such that the little box does not move with respect to the big box
2) Find the tension in the rope in terms of m₁, m₂, μ(k), μ(s), and for this max acceleration
3) Evaluate the tension in the rope at the max acceleration when m₁= 1kg , m₂= 1.5kg , μ(k)= 0.25, μ(s)=0.6 and Ѳ=15

Answer 1

1) maximum acceleration
m1a = m1μ(s)g
a = μ(s)g
2) tension in the rope
Tcosθ = [(m1 + m2)g - Tsinθ]μ(k) + (m1 + m2)μ(s)g
Tcosθ = (m1 + m2)μ(k)g - Tμ(k)sinθ + (m1 + m2)μ(s)g
T(cosθ + μ(k)sinθ) = (m1 + m2)(μ(k)g + μ(s))g
T = (m1 + m2)(μ(k) + μ(s))g/(cosθ + μ(k)sinθ)
3)
T = (1 + 1.5)(0.25 + 0.6)(9.80665)/(cos15° + 0.25sin15°)

Answer 2

This is what I remember.......grain of salt

What ever force exceeds the normal force on the little block will cause it to move.

Fmax=m1*g*(ms)

trig work for max force on lower block
F*cos(angle)=total mass*acceleration-Normal force*(mk)
F*cos(angle) =(m1+m2)*a- (m1+m2)g(mk)
F=(m1+m2)(a-g(mk))/(cos(angle))

so m1g(ms)=(m1+m2)(a-g(mk))/(cos(angle)

a=[m1g(ms)*cos(angle)]/(m1+m2) +g(mk)

a=1(9.81)(.6)cos15/(2.5) + 9.81(.25)

a=2.27+2.45=4.72m/s^2

Answer 3

200mph at 1mln pounds thrust