Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction
between each crate and the surface is 0.43.The crates are pulled to the right at a constant velocity of 2.9 cm/s by a horizontal force. Determine the ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks. Use 9.80 meters per second squared for acceleration due to gravity.
1 answer:
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
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