Find the accelerations a1, a2 , a3 of the three blocks shown in figure. If a horizontal force of 10N is applied on (a) 2 kg block
Answers
(a) When a horizontal force of 10 N is applied to 2 kg block, then the Normal force on 2 kg block will be equal to the weight of 2 kg block
∴ Normal Force = 2g
=20 N
Now, Friction force on 2 kg block = µ₁ × N
=0.2 × 20
= 4 N
Net horizontal force on 2 kg block = 10- 4
= 6 N
Acceleration a₁ =6 N/ 2 kg = 3 m/s²
For 3 kg block,
Upper surface of this block will have frictional force same as above but opposite in direction i.e. 4 N along applied force of 10 N.
Also, The Normal force on the lower surface = Total weight of both blocks
= 5g
= 50 N
Friction force = µ₂ x 50
= 0.3 × 50
= 15 N
But it can not be more than the 4 N. It Further means that the 3 kg and 7 kg blocks do not move with respect to each other.
Now, To calculate the acceleration of these blocks we consider both as if attached. Total mass =3 + 7 = 10 kg.
Total downward force on the smooth surface due to weights = 10g +2g
= 12g
= 120 N
Also, Friction force on this surface = 0 [Since, the friction coefficient of friction is zero].
Now the net external force on this combined block is the friction force applied by the top most block = 4 N.
Acceleration of combined block = Force/mass
= 4/10 kg
= 0.40 m/s²
Hence a₂ = a₃ = 0.40 m/s².
Hope it helps.
(a) When a horizontal force of 10 N is applied to 2 kg block, then the Normal force on 2 kg block will be equal to the weight of 2 kg block
∴ Normal Force = 2g
=20 N
Now, Friction force on 2 kg block = µ₁ × N
=0.2 × 20
= 4 N
Net horizontal force on 2 kg block = 10- 4
= 6 N
Acceleration a₁ =6 N/ 2 kg = 3 m/s²
For 3 kg block,
Upper surface of this block will have frictional force same as above but opposite in direction i.e. 4 N along applied force of 10 N.
Also, The Normal force on the lower surface = Total weight of both blocks
= 5g
= 50 N
Friction force = µ₂ x 50
= 0.3 × 50
= 15 N
But it can not be more than the 4 N. It Further means that the 3 kg and 7 kg blocks do not move with respect to each other.
Now, To calculate the acceleration of these blocks we consider both as if attached. Total mass =3 + 7 = 10 kg.
Total downward force on the smooth surface due to weights = 10g +2g
= 12g
= 120 N
Also, Friction force on this surface = 0 [Since, the friction coefficient of friction is zero].
Now the net external force on this combined block is the friction force applied by the top most block = 4 N.
Acceleration of combined block = Force/mass
= 4/10 kg
= 0.40 m/s²
Hence a₂ = a₃ = 0.40 m/s².
Hope it helps.