Question

A 70-kg man pushes a 50-kg man by a force of 50N. By what force has the 50-kg man pushed the other man?

Answer 1

it is really interesting question .
everyone think that here we have to use $\textbf{Newton's 3r law}$ e.g., action and reaction are of equal in magnitude but directed in opposite direction.
this way answer should be 50N.
means, 50kg man pushed the other man by 50 N force .but it's wrong .

Here we have to use $\textbf{friction concept }$.
here 70kg person pushes a 50kg man by a force of 50N .
Let F = 50N
now, normal reaction act on 50 kg man = mg = 500N { see figure }
we also know, friction is directly proportional to normal reaction then,
fr = kN , k is coefficient of static friction and N is normal reaction.
now, for balancing fr = F
so, 50 = kN = k × 500
k = 0.1 , hence coefficient of static Friction = 0.1

now, Let F' force act by 50kg man on 70 kg man
then, F' = static Friction act on 70kg man
= coefficient of static friction × normal reaction act on 70 kg man
= 0.1 × 70 × 10 [ g = 10 m/s² ]
= 70N

hence, answer should be 70N

Answer 2

its this question Solve by using $\mathit{friction \:\:concept }$.


GIVEN,

70kg person pushes a 50kg man by a force of 50N .


So, Let force = 50N

now, normal reaction act on 50 kg man = mg

mg = 500N

we know, friction is directly proportional to normal reaction then,

fr = kN ,
k is coefficient of static friction and N is normal reaction.

now, for balancing fr = F

so, 50 = kN

50 = k × 500

k = 50/500

K = 0.1 ,

hence coefficient of static Friction = 0.1

Let F' force act by 50kg man on 70 kg man

then, F' = static Friction act on 70kg man

= coefficient of static friction × normal reaction act on 70 kg man

= 0.1 × 70 × 10 [ g = 10 m/s² ]

= 70N