Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 6)
6.
A body A of mass 6.6 kg which is lying on a horizontal platform 4.5 m from its edge is connected to the end of a light string whose other end is supporting a body of mass 3.2 kg as shown in below figure. If the friction between the platform and the body A is 1/3, the acceleration is
Discussion:
6 comments Page 1 of 1.
Mahmood Zaman said:
2 years ago
@SAYYAD.
Take m1=3.2, m2 = 6.6.
Then the answer will be 1m/sec2.
Take m1=3.2, m2 = 6.6.
Then the answer will be 1m/sec2.
(1)
Anagha said:
3 years ago
We need to use D'Alembert's principle for 2blocks.
Assumimg a as acceleration for hanging block we can write Eqn 1 as 3.2g-T=3.2a.
And Eqn 2 as T-(1/3. 6.6g) = 6.6a.
Solving both we get a = 1m/s2.
Assumimg a as acceleration for hanging block we can write Eqn 1 as 3.2g-T=3.2a.
And Eqn 2 as T-(1/3. 6.6g) = 6.6a.
Solving both we get a = 1m/s2.
SAYYAD said:
4 years ago
a=g(m1- μm2)/(m1+m2)
So, m1= 6.6kg.
m2=3.2 kg.
g= 9.81 m/sec^2.
μ = 1/3.
So, by putting these values in the above equation I will get;
a= 5.53 m/sec^2.
Then how you have find this value of
a= 1m/sec^2? Please anyone explain me.
So, m1= 6.6kg.
m2=3.2 kg.
g= 9.81 m/sec^2.
μ = 1/3.
So, by putting these values in the above equation I will get;
a= 5.53 m/sec^2.
Then how you have find this value of
a= 1m/sec^2? Please anyone explain me.
Akki said:
5 years ago
Anyone explain it please.
Swathi s. K said:
7 years ago
a=g(m1-m2μ)/(m1+m2).
Where μ is coefficient of friction,
m1=3.2kg.
m2=6.6kg.
μ=1/3.
And will come 1m/sec^2.
Where μ is coefficient of friction,
m1=3.2kg.
m2=6.6kg.
μ=1/3.
And will come 1m/sec^2.
Kishor said:
8 years ago
Explain, please.
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