static friction\na person is trying to push a box along the floor, but it is not moving. which of the…

static friction\na person is trying to push a box along the floor, but it is not moving. which of the following relationships are correct. (equations are magnitudes only)\n$f_s$ is the static friction on the box from the floor\n$f$ is the force applied by the person to the box\n$f_2$ is on the person from the box\n$f_1$ is on the floor from the box\n$(a = 0)$\na. $f = f_2$ $f\neq f_s$ $f = f_1$\nb. $f>f_s$ and $f_2>f$\nc. $f_2 = f_1$ and $f>f_s$\nd. $f>f_2$ $f = f_s$\ne. $f = f_2$ $f = f_s$ $f_s = f_1$\nselect all
Answer
Explanation:
Step1: Apply Newton's third - law
According to Newton's third - law, the force $F$ applied by the person to the box and the force $F_2$ on the person from the box are an action - reaction pair. So $F = F_2$.
Step2: Analyze the equilibrium of the box
Since the box is not moving ($a = 0$), the net force on the box is zero. The force applied by the person $F$ and the static friction force $f_s$ on the box from the floor balance each other, so $F=f_s$.
Step3: Analyze the force on the floor
The force $F_1$ on the floor from the box and the static friction force $f_s$ on the box from the floor are also an action - reaction pair. So $f_s = F_1$.
Answer:
E. $F = F_2$, $F=f_s$, $f_s=F_1$