Question

static friction
a 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)
$f_s$ is the static friction on the box from the floor
$f$ is the force applied by the person to the box
$f_2$ is on the person from the box
$f_1$ is on the floor from the box
$(a = 0)$
a. $f = f_2$ $f
eq f_s$ $f = f_1$
b. $f>f_s$ and $f_2>f$
c. $f_2 = f_1$ and $f>f_s$
d. $f>f_2$ $f = f_s$
e. $f = f_2$ $f = f_s$ $f_s = f_1$
select all

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. $F = F_2$, $F=f_s$, $f_s=F_1$