Thank you for registering.

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

Please check your email for login details.
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 bloack of mass under root 3kg is resting on a horizontal plane (coefficient of static friction =1/2under root3 ) . A foce is applied to the block at an angle of 60 . The minimum magnitude of force for which the block begins to slide is (g =10m/s square) please explain

a bloack of mass under root 3kg is resting on a horizontal plane (coefficient of static friction =1/2under root3 ) . A foce is applied to the block at an angle of 60 . The minimum magnitude of force for which the block begins to slide is (g =10m/s square)      please explain

Grade:11

3 Answers

vikas askiitian expert
509 Points
10 years ago

taking two components of force seperatly fcos@ and f sin@  

 fcos@ will become driving force and

  (mg-fsin@) will be the normal reaction acting on the body

  we know, friction force= u.normal reaction ,u=cofficient of friction

for minimum force fcos@=friction force 

    f=20/3 N

 

Xyz
13 Points
4 years ago
Normal reaction:Fsin60+mg
Root3(F/2+10). 
As frictional force= u.N
u=coefficient of friction),N = normal reaction
 
Fcos60>(or equal)(F+20)/4
On solving, min F =20N
 
SIDDHARTH
15 Points
3 years ago
F=20N Ans
Frictional Force = Fcos60 (begins slide)
u.n= F* ½
1/2root3*(mg+F sin60) = F*1/2
F/4 + 5 = F*1/2
Hence F= 20 N
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free