Discussion :: Friction - General Questions (Q.No.4)
|Shiva said: (Nov 21, 2011)|
|How could it be? It is 80lb Tips|
|Svenu said: (Sep 7, 2012)|
|It could 80 lb why 75 can any one tell me|
|Naveen Goud said: (Mar 13, 2013)|
|Take the moments of p&weight about other end of bottom point then p*4+(200*1.5)=0 p=75 &tips about that point.|
|Adla said: (Jul 18, 2014)|
|Can anyone give me the working out for this question?|
|Qahtan said: (Jul 1, 2015)|
|tan (theta) = (meu_k) = 1.5/h; (meu_k) = 0.4.
0.4 = 1.5/h ; h = 3.75.
Where h = 4 (meu_k) = 1.5/4 = 0.375.
Then at (meu_k) = 0.4 the refrigeration is tips.
And at (meu_k) = 0.375 the refrigeration is slips.
Sum moments about other end equal to zero.
P*4-(200*1.5) = 0; p = 200*1.5/4; p = 75 lb tips.
|Shashi Kumar said: (Nov 3, 2016)|
|Please, tell me the right answer with proper explanation.|
|Arun Kumar said: (Jan 4, 2017)|
|Answer C is also possible when we take equilibrium condition. As we considering friction at the base & normal reaction. Its also correct.|
|Prahlad said: (Oct 2, 2017)|
|It means p is the value in which body are statics or move then the answer is D.|
|Prahlad Bhardwaj said: (Oct 2, 2017)|
|P =200lb*1.5/4 =75ans.|
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