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.

P = 75 lb Slips
P = 80 lb Tips
P = 80 lb Slips
P = 75 lb Tips
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

JAYA DEVA CHARYA said:   3 years ago
In this case, the maximum static friction value is = 0.4*200 = 80lb (Force required to move the body in a linear direction or slide the body)

Taking moment about at the tipping position,
P*4 = 1.5*200
P = 75lb ( Tipping condition).

So, since the value of tipping is less it will occur first the body will tip first with 75lb of force.

Aaqib said:   3 years ago
I think the answer is C.

Prahlad bhardwaj said:   8 years ago
P =200lb*1.5/4 =75ans.

Prahlad said:   8 years ago
It means p is the value in which body are statics or move then the answer is D.

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.

Shashi kumar said:   9 years ago
Please, tell me the right answer with proper explanation.

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.

Adla said:   1 decade ago
Can anyone give me the working out for this question?

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.

Svenu said:   1 decade ago
It could 80 lb why 75 can any one tell me


Post your comments here:

Your comments will be displayed after verification.