Question

1. add in the missing forces so the box will move to the left with a net force of 45n.

Explanation:

Step1: Define net force formula

Let the left - ward force be \( F_{left} \), the right - ward force be \( F_{right}=65N \), and the net force \( F_{net} = 45N \) (left - ward). The formula for net force in one - dimensional motion (horizontal direction here) is \( F_{net}=F_{left}-F_{right} \) (since left - ward is positive).

Step2: Solve for left - ward force

We know \( F_{net} = 45N \) and \( F_{right}=65N \). From \( F_{net}=F_{left}-F_{right} \), we can re - arrange the formula to \( F_{left}=F_{net}+F_{right} \). Substituting the values, we get \( F_{left}=45N + 65N=110N \).

For the vertical direction (second diagram), let the upward force be \( F_{up}=10N \), and let the total downward force be \( F_{down} \). For the object to be in vertical equilibrium (since there is no net vertical motion mentioned, we assume vertical forces are balanced), so \( F_{down}=F_{up} = 10N \). But if we consider the two downward arrows, we can assume they are equal (if we need to find the magnitude of each downward force, let each downward force be \( F_{1}\) and \( F_{2}\), and \( F_{1}+F_{2}=10N \). If we assume they are equal, then \( F_{1}=F_{2} = 5N \) each. But the main part of the problem is about the horizontal force for the first box.

Answer:

The left - ward force should be \( 110N \), and in the vertical direction, if the two downward forces are equal, each is \( 5N \). (For the horizontal net force part, the left - ward force is \( 110N \))