Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is µ. Find the acceleration of the system and the force by the rod on one of the blocks..Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "
Answers
Answered by
22
Solution :
From the free body diagram
R1=M1g cos θ-----------1
R2=M2g Cosθ-------------2
T+M1gsinθ-M1a-μ1R1=0 ------------3
T-M2g+M2a+μR=0---------4
From equation 3
T+M1g Sinθ-M1a-cosθ=0 ------------------5
From equation v
gsinθ(M1+M2) -μgcosθ(M1+M2) --------------7
a(M1+M2)=g sinθ(M1+M2)-μgCosθ(M1+M2)
a=g (Sinθ-μcosθ)
The block has acceleration =g(sinθ-μcosθ)
The force exerted by the rod on one of the blocks is tension
Tension T=-M1gSinθ+M1(gsinθ-μgcosθ)+μM1gcosθ
Tension T=0
From the free body diagram
R1=M1g cos θ-----------1
R2=M2g Cosθ-------------2
T+M1gsinθ-M1a-μ1R1=0 ------------3
T-M2g+M2a+μR=0---------4
From equation 3
T+M1g Sinθ-M1a-cosθ=0 ------------------5
From equation v
gsinθ(M1+M2) -μgcosθ(M1+M2) --------------7
a(M1+M2)=g sinθ(M1+M2)-μgCosθ(M1+M2)
a=g (Sinθ-μcosθ)
The block has acceleration =g(sinθ-μcosθ)
The force exerted by the rod on one of the blocks is tension
Tension T=-M1gSinθ+M1(gsinθ-μgcosθ)+μM1gcosθ
Tension T=0
Answered by
1
Answer:
Hey guys
See the attachment below....❤x
Attachments:
Similar questions