Answer:
T1=5.588N
T2=4.543N
Explanation:
A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of and a radius of 15.0 cm. A mass of 710 g hangs from the other end of the string. When the tiles are released, the larger one accelerates downward while the lighter one accelerates upward. The pulley has no friction in its axle and turns without the string slipping. What is the tension in the string on the side of the 710-g tile?
convert grams to kilograms
Let
M₁ = mass 1 = 0.710 kg
M₂ = mass 2 = 0.385 kg
I = moment of inertia of the pulley = 0.0125 kgm²
R = radius of the pulley = 0.15 m
T₁ = tension in the string in connection to M₁
T₂ = tension in the string in connection to M₂
A = acceleration of the system
lets ave a deep dive into the formula
M₁×g - T₁ = M₁×A
(0.710 kg)×(9.8 m/s²) - T₁ = (710 kg)×A
(6.958 N) - T₁ = (0.710 kg)×A
T₁ = (6.958N) - (0.710 kg)×A . . . . . . . equation 1
second equation: as follows
T₂ - M₂×g = M₂×A
T₂ - (0.385 kg)×(9.8 m/s²) = (0.385 kg)×A
T₂ - (3.773N) = (0.385 kg)×A
T₂ = (0.385 kg)×A + (3.773 N) . . . . . . . 2
third equation:
T₁×R - T₂×R = I×A/R
T₁×(0.15 m) - T₂×(0.15 m) = (0.0125 kgm²)×A/(0.15 m)
T₁×(0.15 m) - T₂×(0.15 m) = (0.0833 kgm)×A . . . . . . . 3
substitute the values f 1 and 2 in equation 3
T₁ = (6.958N) - (0.710 kg)×A
T₂ = (0.385 kg)×A + (3.773 N)
(6.958N) - (0.710 kg×A)×(0.15 m) - (0.385 kg)×A + (3.773 N)×(0.15 m) = (0.0833 )×A . . . . . . . 3
1.0437-0.106A-(0.05A+.565)=0.0833A
0.477=0.0833A+0.05A+0.106A
0.477=0.2393A
A=1.993m^2
A=2m/s^2
from equation 1
T₁ = (6.958N) - (0.710 kg)×2
T1=5.588N
from equation 2
substituting the value of A, which is the acceleration
T₂ = (0.385 kg)×2 + (3.773 N)
T2=4.543N
