Question

Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is µ1, and that between the block of mass 4.0 kg and the incline is µ2. Calculate the acceleration of 2.0 kg block if (a) µ1 = 0.20 and µ2= 0.30 , (b) µ1 = 0.30 and µ2=0.20. Take g = 10 m/s².Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "

Answer 1

Solution :
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From the free body diagram :
R=4gcosθ
R=4x10√3/2=20√3 ---------------(1)
μ2R+4a-p-4gsin30°=0
0.3x40 cos 30°+4a-p-40sin20°=0-------------(2)
R1=2gcos30°=10√3---------(3)
p+2a+μ1R1-2gsin30°=0----------(4)
From equation , (ii) 6√3+4a-p-20=0
from equation iv, p+2a+2√3-10=0/
6√3+6a=30+2√3=0
6a=30-8√3=30-13.85
=16.15
a=16.15/6=2.69=2.7m/s2

b) In this case , 4kg block will travel with more acceleration because coefficient of friction is less than that of 2kg . so, they will move separately.
From the free body diagram of 2kg mass, we can find out that a=2.4m/s2

Answer 2

Answer:

From the free body diagram, we get

T+0.5a-0.5g=0-----------------(1)

μR +1a+T1 -T=0-----------------(2)

μR+1a-T1=0

μR +a= T1------------(3)

From equations i , ii and iii

μR+a=T-T1

T-T1=T1

T=2T1

Equation II becomes μR +a+T1-2T1=0

μR+a-T1=0

T1=μR+a

T1=0.2g +a

Equation I will become  

2T1+0.5a-0.5g =0

T1=0.5g -0.5a/2  

=0.25g-0.25a

From equation Iv and V

0.2g +a=0.25g -0.25 a

a=0.05/1.25 x10  

a=0.4 x 10

a=0.4 m/s2

Therefore acceleration of 1kg block is 0.4m/s2

b) Tension T1=0.2g +a+0.4=2.4 N

c) t=0.5g-0.5a

t=0.5x10-0.5x0l4

t=4.8N