Chapter 5, Problem 59A

To determine

To rank: The magnitude of the net force from least to greatest.

Answer to Problem 59A

From least to greatest the magnitude of net force is A (10 N), D (19.1 N), E (20 N), B (22.4 N), C (30)

Explanation of Solution

Given:

The given forces are:

1. 20 N up + 10 N down.
2. 20 N up + 10 N left.
3. 20 N up +10 N up.
4. 20 N up + 10 N $20°$ below the horizontal.
5. 20 N up.

Formula Used:

The resultant (net force) of two forces F₁ and F₂ are given by the formula

  $F_{net}=\sqrt{F_1{}^2+F_2{}^2+2F_1{F}_2\cos \theta }$

Calculation:

The resultance for all the cases is,

For (A), 20 N up +10 N down

F₁ = 20N

F₂ = 10 N

Consider the force diagram shown below.

The angle between F₁ and F₂ is $\theta =180°$ .Therefore the net force is

  $\begin{array}{l}{F}_{net}\text{= }\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}\text{+}\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\text{cos180}°\right)}\\ F_{net}\text{= }\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}-\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\right)}\\ F_{net}=\sqrt{{\left(20\text{N}-\text{10N}\right)}^2}\\ F_{net}\text{= }\left(\text{20N}\right)-\left(\text{10N}\right)\\ F_{net}\text{= 10 N}\end{array}$

For (B), 20 N up +10 N left

F₁ = 20 N

F₂ = 10 N

Consider the force diagram shown below.

The angle F₁ and F₂ is $\theta =90°$ . The net force is

  $\begin{array}{l}{F}_{net}=\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}\text{+}\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\text{cos90}°\right)}\\ F_{net}=\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}}\\ F_{net}=\text{ 22}\text{.36 N}\end{array}$

For (C), 20 N up + 10 N up

F₁ = 20 N

F₂ = 10 N

The angle betweenF₁ and F₂ is $\theta =0°$ . The net force is

  $\begin{array}{l}{F}_{net}\text{= }\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}\text{+}\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\text{cos0}°\right)}\\ F_{net}\text{= }\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}\text{+}\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\right)}\\ F_{net}\text{= }\left(\text{20N}\right)\text{+}\left(\text{10N}\right)\\ F_{net}\text{= 30 N}\end{array}$

For (D), 20 N up +10 N $20°$ below the horizontal.

F₁ = 20 N F₂ = 10 N Consider the force diagram as shown below.

The angle F₁ and F₂ is $\theta =110°$ . The net force is

  $\begin{array}{l}{F}_{net}=\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}\text{+}\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\cos 110°\right)}\\ F_{net}=\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}\text{+}\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\left(-0.342\right)\right)}\\ F_{net}=\sqrt{{\left(\text{20N}\right)}^{\text{2}}\text{+}{\left(\text{10N}\right)}^{\text{2}}-\left(\text{2}\left(\text{10N}\right)\left(\text{20N}\right)\left(0.342\right)\right)}\\ F_{net}=\text{19}\text{.06 N}\end{array}$

For (E), 20N up

Since there is only one force, so net force will be $F_{net} =20\text{ N}$

Conclusion:

From the above calculation ,the rank of net force from least to high is

1.A (10 N)

2.D (19.06 N)

3.E (20 N)

4.B (22.36 N)

5.C (30 N)