Engineering Mechanics - Friction - Discussion

Discussion :: Friction - General Questions (Q.No.4)

4. 

The refrigerator has a weight of 200 lb and a center of gravity at G. Determine the force P required to move it. Will the refrigerator tip or slip? Take = 0.4.

[A]. P = 75 lb Slips
[B]. P = 80 lb Tips
[C]. P = 80 lb Slips
[D]. P = 75 lb Tips

Answer: Option D

Explanation:

No answer description available for this question.

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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