Engineering Mechanics - Friction - Discussion
Discussion Forum : 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.
Discussion:
11 comments Page 1 of 2.
Shiva said:
1 decade ago
How could it be? It is 80lb Tips
Svenu said:
1 decade ago
It could 80 lb why 75 can any one tell me
Naveen Goud said:
1 decade ago
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:
1 decade ago
Can anyone give me the working out for this question?
Qahtan said:
1 decade ago
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.
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:
9 years ago
Please, tell me the right answer with proper explanation.
ARUN KUMAR said:
9 years ago
Answer C is also possible when we take equilibrium condition. As we considering friction at the base & normal reaction. Its also correct.
Prahlad said:
8 years ago
It means p is the value in which body are statics or move then the answer is D.
Prahlad bhardwaj said:
8 years ago
P =200lb*1.5/4 =75ans.
Aaqib said:
3 years ago
I think the answer is C.
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