×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A body of mass 5kg is placed on rough horizontal surface. If coefficient of friction is 1/√3, find how much pulling force should act on the body at an angle 30° from horizontal so that the body just begins to move.


3 years ago

## Answers : (2)

Arun
24742 Points
							n = mg - Fmax/2n = Fmax/(1/$\small \sqrt$3)n = $\small \sqrt$3 FmaxThusmg - Fmax/2 = $\small \sqrt$3 FmaxKnowing thatmg = 5*9.8 = 49 Hence Fmax = 22 N

3 years ago
Samar
13 Points
							we can also find  fmax  directly by rearranging the two blue equations to solve for  n , and then setting them equal to each other: n=mg−fmax2  n=fmax1√3=√3fmax  Thus, mg−fmax2=√3fmax  Knowing that  mg=(5lkg)(9.81lm/s2) , fmax=22.0   N

2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Modern Physics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions