A system consists of 3 objects M1, M2 and M3 as shown in the diagram below. Consider the coefficient of kinetic friction between the block and surface to be 0.256. Calculate,
i. The friction force on the system
ii. The normal force
iii. The acceleration of the system
iv. In which direction will the system move
Answers
Answer:
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Given: Three objects with mass M1, M2, and M3 respectively.
K.E. Friction Coefficient between the block and surface= 0.256
To find: (i) The friction force on the system.
(ii) The normal force
(iii) The acceleration of the system
(iv) In which direction will the system move.
Solution:
To calculate the friction force, we need to find the normal force first.
(ii) The normal force
The two objects ( ma and m3) are at rest then
Fₙ = (m2 +m3)x g
= (m2+m3)x 10 // gravity constant as 10 and if given use that value only
= 10(m2+m3)
Since F(gravity) = - F(normal)
F(normal) = - 10(m2+m3)N
Since the weight act downwards in case of m1
0N = (-10N) + Fn
F(normal) = 10 N
F(net) = F(m2+m3) + F(m1)
= -10(m2+m3)N + 10N
= 10( -(m2 +m3)) N
The normal force will be 10( m2 +m3) N
(i) The friction force on the system
F(friction) = K.E coef. x F(Normal)
= 0.256 x 10( -(m2 +m3)) N
= 2.56(-(m2+m3))N
Friction is always in opposite direction
F(friction)= 2.56(m2+m3)N
The friction force on the system is 2.56(m2+m3)N
(iii) The acceleration of the system
The motion of the horizontal block i.e. m2 and m3 = (m2 +m3) x a = T- f₂ - f₃
or you can write it as (m + m)a = T - µmg - µmg
or 2ma = T-2 µmg -(eq 1)
For the motion of vertical block m₁a= m₁g - T or ma = mg-T - (eq 2)
From eq 1 and 2
2ma = mg-ma-2 µmg
2ma +ma = mg - 2 µmg
3 ma = mg - 2 µmg
3 ma = m( g - 2 µg)
Eliminating m from both sides
3a = g - 2 µg
a = g - 2 µg / 3
Acceleration of the system is g - 2 µg / 3
(iv) Direction in which the system will move
The system always moves opposite to the friction direction since friction is positive, friction will be in the forward direction then the system will move in the backward direction.
