Answer:
R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks
Explanation:
We apply Newton's second law:
∑F=m*a
velocity is constant ,then , a=0
Nomenclature
W: weight
m: mass
N : normal force
Ff: Friction force
μk: coefficient of kinetic friction
T: tension force in the rope
F: applied horizontal force
g: acceleration due to gravity.
Force Calculation
W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N
W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N
∑Fy=0
N₁-W₁=0 , N₁=W₁ = 52.92 N
N₂-W₂=0, N₂=W₂=80.36N
Ff₁= μk* N₁=0.4*52.92 N = 21.16N
Ff₂= μk* N₂=0.4*80.36N = 32.14N
Look at the attached graphic
Free-Body diagram m₁=5.4 kg
∑Fx=0
T- Ff₁=0 , T= Ff₁ , T= 21.16N
Free-Body diagram m₂=8.2 kg
∑Fx=0
F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N
Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)
R= F/T= 53.3N/21.16N = 2.5
