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Answers
QUESTION:
Three blocks of masses 6 kg, 9 kg, 10 kg are attached as shown in Figure. Coefficient of friction between the table and the 10 kg block is 0.2. What is the acceleration of the system? What are the tensions in the two strings?
ANSWER:
- The acceleration of the system = 0.392 m/s².
- Tension in the string of 9 kg block is 84.67 N
- Tension in the string of 6 kg block is 61.15 N
GIVEN:
- Three blocks of masses 6 kg, 9 kg, 10 kg are attached.
- Coefficient of friction between the table and the 10 kg block is 0.2.
TO FIND:
- The acceleration of the system.
- The tensions in the two strings.
EXPLANATION:
Let the tension in the 9 kg block be T₁
9 kg block accelerates downwards.
By free body diagram,
mg - T₁ = ma
m = 9 kg
9g - T₁ = 9a
T₁ = 9g - 9a
Let the tension in the 6 kg block be T₂
6 kg block accelerates upwards.
By free body diagram,
T₂ - mg = ma
m = 6 kg
T₂ - 6g = 6a
T₂ = 6g + 6a
10 kg block accelerates towards right.
By free body diagram,
T₁ - T₂ - μmg = 10a
μ = 0.2
m = 10 kg
T₁ = 9g - 9a
T₂ = 6g + 6a
9g - 9a - 6g - 6a - 0.2(10)g = 10a
3g - 15a - 2g = 10a
g = 10a + 15a
25a = g
g = 9.8 m/s²
a = 9.8/25
a = 0.392 m/s²
Hene the acceleration of the system = 0.392 m/s².
T₁ = 9g - 9a
a = 0.392 m/s²
g = 9.8 m/s²
T₁ = 9(9.8 - 0.392)
T₁ = 9(9.408)
T₁ = 84.67 N
T₂ = 6g + 6a
a = 0.392 m/s²
g = 9.8 m/s²
T₂ = 6(9.8 + 0.392)
T₂ = 6(10.192)
T₂ = 61.15 N
Hence tension in the string of 9 kg block is 84.67 N and tension in the string of 6 kg block is 61.15 N.
