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## Homework Statement

Box 1 of mass m1 is attached by a thin rope to box 2 of mass m2. Box 1 is pulled by force F. What is the acceleration of box 1 given:

m1= 3kg

m2= 2kg

V(initial)= 4m/s

F= 50N

Friction is present between boxes and floor

μ = 0.2

θ = 20 degrees

∅ = 35 degrees

gravity(g) = 9.8

And what value would F need to be so the velocity is constant at 4.0 m/s

## The Attempt at a Solution

I know that Fnet is going to be the x component of the force F minus the friction between box 1 and the ground and the x component of the tension on box 1 by box 2

Fnet = Fx-(Fμ+Ftx)

By plugging into equations I found

Fx = cos∅*F = 40.9576

Fy = sin∅*F = 28.6788

Fg = m*g = 29.4

N(normal) = Fg+ Fy+Fty=50.0788+Fty

Fμ = N*μ=μ(Fg+Fy+Fty)=0.2(50.0788+Fty)

Fnet =Fx-(Fμ+Ftx)=40.9576-(0.2(50.0788+Fty)+Ftx)

I just can't find Ft

I initially tried that Fty=m2gsinθ and Ftx=m2gcosθ bur this was marked wrong