Question

(a) Find the magnitudes of the forces F₁ and F₂ that add to give the total force F_tot shown in Figure 4.35. This may be done either graphically or by using trigonometry.

(b) Show graphically that the same total force is obtained independent of the order of addition of F₁and F₂ .

(c) Find the direction and magnitude of some other pair of vectors that add to give F_tot . Draw these to scale on the same drawing used in part (b) or a similar picture.

Short answer

(a) The magnitude of force F₁ is 16.38 N, and the magnitude of force F₂ is 11.47 N.

(c) the magnitude of F₁’ is 17.43 N and the magnitude of F₂’ is 9.80 N.

Step-by-step solution

(a) Determine the magnitude of the forces F1 and F2

Since F₁ is the x-component of the total force and F₂ is the y-component of the total force, therefore,

${\mathit{F}}_{\mathbf{1}}\mathbf{=}{\mathit{F}}_{\mathbf{t}\mathbf{o}\mathbf{t}}\mathbf{cos}\mathit{\theta }$………………. (i)

${\mathit{F}}_{\mathbf{2}}\mathbf{=}{\mathit{F}}_{\mathbf{t}\mathbf{o}\mathbf{t}}\mathbf{sin}\mathit{\theta }$………………. (ii)

Here, F₁ is the horizontal force, F_tot is the total force, $\mathit{\theta }$is the angle between the total force and the horizontal force, and F₂ is the vertical force.

Substitute 20 N for F_tot, and$35°$for$\theta$in equation (i), and we get,

$\begin{array}{c}{F}_1=20\text{\hspace{0.33em}N}\times \cos {35}^{°}\\ =20\text{\hspace{0.33em}N}\times 0.819\\ =16.38\text{\hspace{0.33em}N}\end{array}$

Substitute 20 N for F_tot, and$35°$for$\theta$in equation (ii), and we get,

$\begin{array}{c}{F}_2=20\text{\hspace{0.33em}N}\times \sin {35}^{°}\\ =20\text{\hspace{0.33em}N}\times \text{0.57357}\\ =\text{11.47\hspace{0.33em}N}\end{array}$

Hence, the magnitude of force F₁ is 16.38 N, and the magnitude of force F₂ is 11.47 N.

(b) Graphical representation by changing the order of addition of F1 and F2

Here, figure (a) is the same as figure (b).

Hence, the same total force is obtained by changing the order of F1 and F₂.

(c) Graphical representation by changing the order of addition of F1 and F2

Let F₁’ be the force at an angle of $20°$with the horizontal. Then, calculate its magnitude.

$F_1\text{'}=\frac{F_1}{\cos {20}^{°}}$

Substitute 16.38 N for F₁.

$\begin{array}{c}{F}_1\text{'}=\frac{16.38\text{\hspace{0.33em}N}}{\cos {20}^{°}}\\ =\frac{16.38}{0.9397}\\ =17.43\text{\hspace{0.33em}N}\end{array}$

And F₂’ is at an angle of 90̊ from the horizontal. Then, calculate its magnitude.

$F_2\text{'}=\sqrt{{\left(F_{total}\right)}^2-{\left(F_1\text{'}\right)}^2}$

Substitute 20 N for F_total and 17.43 N for F₁’.

$\begin{array}{c}{F}_2\text{'}=\sqrt{{\left(20\text{\hspace{0.33em}N}\right)}^2-{\left(17.43\text{\hspace{0.33em}N}\right)}^2}\\ =9.80\text{\hspace{0.33em}N}\end{array}$

Hence, the magnitude of F₁’ is 17.43 N and the magnitude of F₂’ is 9.80 N.