Find the acceleration of the system and the tension T1 and T2 where the coefficient of friction between the block and table is 0.2: m1=2kg (left side downward pull), m2=5kg (center block on table), m3=10kg (right side downward pull. T1 between m1/m2; T2 between m2/m3.
Answers
Solution :
⏭ Given:
✏ The co-efficient of friction between the block and table = 0.2
✏ Tension b/w m1 and m2 = T1
✏ Tension b/w m2 and m3 = T2
✏ mass of m1 = 2kg
✏ mass of m2 = 5kg
✏ mass of m3 = 10kg
⏭ To Find:
- Acceleration of system
- Tension forces (T1 and T2)
⏭ Concept:
✏ First, we have to find out acceleration of system with the help of net force concept after that we can calculate tension force in the string with the help of F.B.D. of any two blocks.
⏭ Calculation:
_________________________________
- Acceleration of system :
✏ Fnet = (m1 + m2 + m3) × a
✏ m3g - m2g -m1g = 17a
✏ (10×10)-(0.2×5×10)-(2×10) = 17a
✏ 100 - 10 - 20 = 17a
✏ 70 = 17a
✏ a = 4.12 m
_________________________________
- Tension force T1 :
✏ Net force on 2kg mass...
✏ T1 - m1g = m1a
✏ T1 - 20 = 2 × 4.12
✏ T1 = 8.24 + 20
✏ T1 = 28.24 N
_________________________________
- Tension force T2 :
✏ Net force on 10kg mass...
✏ m3g - T2 = m3a
✏ 100 - T2 = 10 × 4.12
✏ T2 = 100 - 41.2
✏ T2 = 58.8 N
_________________________________
Given:
Pulley , blocks system has been provided. Coefficient of Friction between block and table is 0.2 .
To find:
Acceleration of blocks, T1 and T2
Concept:
First draw the FBD (Free Body diagram ) of all the blocks :
For 10 kg Block :
10g - T2 = 10a .........(1)
For 5 kg Block:
T2 - T1 - μN = 5a
=> T2 - T1 - 5g(0.2) = 5a
=> T2 - T1 - 10 = 5a ........(2)
For 2 kg Block :
T1 - 2g = 2a ........(3)
Adding all the Equations : (Putting g=10 m/s²)
100 - 10 - 20 = 10a + 5a + 2a = 17a
=> 17a = 70
=> a = 70/17
=> a = 4.11 m/s²
Putting value of acceleration in the above Equation:
10g - T2 = 10a
=> 100 - T2 = 41.1
=> T2 = 100 - 41.1
=> T2 = 58.9 N
Again:
T1 - 2g = 2a
=> T1 - 20 = 2(4.11)
=> T1 = 8.22 + 20
=> T1 = 28.22 N