Question

2. Free-body diagrams for four situations are shown below. The net force is known for each situation.However, the magnitudes of a few of the individual forces are not known. Analyze cach situationindividually and determine the magnitude of the unknown forces.TB30050 l80 Nr20 NG200200 kFnet = 0 NFnet- 900 N, upet = 60 N, leftFnet = 30 N, rightClick to add speaker notes

Answer 1

In order to calculate the missing forces, we need to know that the net force is the sum of all forces in a direction.

For the first situation, the net force is zero, so the sum of forces in each direction is zero.

In the vertical direction, we have:

`\begin{gathered} F_(net)=B-200\\ \\ 0=B-200\\ \\ B=200\text{ N} \end{gathered}`

In the horizontal direction:

`\begin{gathered} F_(net)=50-A\\ \\ 0=50-A\\ \\ A=50\text{ N} \end{gathered}`

For the second situation, the net force is 900 N up.

In the vertical direction, we have:

`\begin{gathered} F_(net)=C-200\\ \\ 900=C-200\\ \\ C=1100\text{ N} \end{gathered}`

In the horizontal direction, we have no forces.

For the third situation, the net force is 60 N to the left.

In the vertical direction, we have:

`\begin{gathered} F_(net)=300-E\\ \\ 0=300-E\\ \\ E=300\text{ N} \end{gathered}`

In the horizontal direction, we have:

`\begin{gathered} F_(net)=D-80\\ \\ -60=D-80\\ \\ D=20\text{ N} \end{gathered}`

For the fourth situation, the net force is 30 N to the right.

In the vertical direction, we have:

`\begin{gathered} F_(net)=F-H\\ \\ 0=F-H\\ \\ F=H \end{gathered}`

In the horizontal direction, we have:

`\begin{gathered} F_(net)=G-20\\ \\ 30=G-20\\ \\ G=50\text{ N} \end{gathered}`