Question

a free body diagram is shown below.
if ( f<210 mathrm{~n} ), the ( sum vec{f} ) acting on the object is zero.
if ( f = 210 mathrm{~n} ), the ( sum vec{f} ) acting on the object is.
if ( f>210 mathrm{~n} ), the ( sum vec{f} ) acting on the object is.

Explanation:

Step1: Determine the equilibrium condition

The net force \(\sum\vec{F}\) is zero when the sum of forces to the left equals the sum of forces to the right. Let's assume the right - hand direction is positive. The force to the right is \(300N\), and the forces to the left are \(90N\) and \(F\). So, \(\sum\vec{F}=300-(90 + F)\). When \(\sum\vec{F} = 0\), we have \(300-(90 + F)=0\), which gives \(F = 210N\).

Step2: Analyze when \(F<210N\)

If \(F<210N\), then \(90 + F<300\). So, \(\sum\vec{F}=300-(90 + F)>0\) (but the problem states when \(F < 210N\), the net force is zero? No, wait, let's re - calculate the net force formula correctly. The net force \(\sum\vec{F}\) in the horizontal direction: \(\sum\vec{F}=300-(90 + F)\). When \(F = 210N\), \(\sum\vec{F}=300-(90 + 210)=0\). When \(F<210N\), say \(F = 200N\), \(\sum\vec{F}=300-(90 + 200)=10N\) (wrong, no, wait, the formula for net force: assume right is positive. Forces: \(+ 300N\) (right), \(-90N\) (left), \(-F\) (left). So \(\sum\vec{F}=300-90 - F=210 - F\). When \(F = 210N\), \(\sum\vec{F}=0\). When \(F<210N\), \(\sum\vec{F}=210 - F>0\) (but the problem's third statement says if \(F < 210N\), \(\sum\vec{F}\) is zero. There is a mistake in the problem setup. But following the problem's logic:
The equilibrium condition (net force \(\sum\vec{F}=0\)) gives \(300=90 + F\), so \(F = 210N\).
If \(F>210N\), then \(90+F>300\), and \(\sum\vec{F}=300-(90 + F)\) (left - hand direction, negative value in our sign convention). If \(F = 210N\), \(\sum\vec{F}=0\). If \(F<210N\), \(90 + F<300\), \(\sum\vec{F}=300-(90 + F)>0\) (but the problem says for \(F < 210N\), \(\sum\vec{F}\) is zero. Assuming the problem has a typo and we follow the given fill - in - the - blank:
When \(F>210N\), \(\sum\vec{F}\) is in the left - hand direction (negative if right is positive). When \(F = 210N\), \(\sum\vec{F}=0\). When \(F<210N\), \(\sum\vec{F}\) is in the right - hand direction. But according to the problem's structure:
If \(F>210N\), \(\sum\vec{F}\) is non - zero (leftward). If \(F = 210N\), \(\sum\vec{F}=0\). If \(F<210N\), \(\sum\vec{F}\) is non - zero (rightward). But the problem's third statement says "if \(F < 210N\), the \(\sum\vec{F}\) acting on the object is zero" which is wrong. But following the problem's given blanks:
The net force formula \(\sum\vec{F}=300-(90 + F)\)
When \(F>210N\), \(\sum\vec{F}=300-(90 + F)<0\) (left - ward, magnitude \(F - 210\))
When \(F = 210N\), \(\sum\vec{F}=300-(90 + 210)=0\)
When \(F<210N\), \(\sum\vec{F}=300-(90 + F)>0\) (right - ward, magnitude \(210 - F\))

Answer:

If \(F>210N\), the \(\sum\vec{F}\) acting on the object is \(-(F - 210)N\) (left - ward). If \(F = 210N\), the \(\sum\vec{F}\) acting on the object is \(0N\). If \(F<210N\), the \(\sum\vec{F}\) acting on the object is \((210 - F)N\) (right - ward). But following the problem's given blanks (assuming some misprint in the problem's third statement):
If \(F>210N\), the \(\sum\vec{F}\) acting on the object is non - zero (left - ward). If \(F = 210N\), the \(\sum\vec{F}\) acting on the object is \(0\). If \(F<210N\), the \(\sum\vec{F}\) acting on the object is non - zero (right - ward). But if we strictly follow the calculation based on \(\sum\vec{F}=300-(90 + F)\):
For \(F>210N\): \(\sum\vec{F}=300-(90 + F)=210 - F<0\) (e.g., if \(F = 220N\), \(\sum\vec{F}=210 - 220=-10N\))
For \(F = 210N\): \(\sum\vec{F}=0\)
For \(F<210N\): \(\sum\vec{F}=210 - F>0\) (e.g., if \(F = 200N\), \(\sum\vec{F}=210 - 200 = 10N\))