Question

A body of weight 20N is on a horizontal surface, minimum force
applied to pull it when applied force makes an angle
0
60
with
horizontal (angle of friction
0  = 30
) is

Answer 1

Tan 30°= coeff of friction

u=1/√3

Min force applied=limiting friction

F cos 60°=1/√3[weight - F sin 60°]

F/2+F/2=weight/√3

F=20/√3 N

Answer 2

Answer: The answer is 20/root over 3N

Explanation: mg=20 theta=30 angle of friction=30

So we are gonna use the formula

mg sin theta/cos ( theta - the angle of friction)

Substituting the following values we get 20(1/2)/cos(60-30)

Which gives us 10/cos 30

cos 30 value is root over 3/2

Substituting Cos 30 value we get 10/root over 3/2

Send the two up 10 time 2 =20

Therefore we get 20/ root over 3N

As our Answer