F=m*a=>a=F/m=10000/1267=7.89 m/s²
d=v₀t+a't²/2<=>394.6=112.5*a'=>a'=3.5
a-a'=7.89-3.5=4.39 m/s²
This difference causes friction forces
We apply the second principle of dynamics: vector: F + N + G + Ff = ma (vector vectors, I can not here)
Scalar: Ox: F-Ff = ma
Oy N-mg = 0
Ff = -ma+ F =-1267*7.89+10000=-8869+10000=1131 N
This frictional force (Ff) is opposite to the traction (F)
First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
The answer is C why because I know it is