91影视

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

${\mathit{F}}_{\mathbf{1}}\mathbf{=}{\mathit{F}}_{\mathbf{t}\mathbf{o}\mathbf{t}}\mathbf{cos}\mathit{胃}$鈥︹�︹�︹�︹�︹��. (i)

${\mathit{F}}_{\mathbf{2}}\mathbf{=}{\mathit{F}}_{\mathbf{t}\mathbf{o}\mathbf{t}}\mathbf{sin}\mathit{胃}$鈥︹�︹�︹�︹�︹��. (ii)

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

Substitute 20 N for F_tot, and$35掳$for$胃$in equation (i), and we get,

$\begin{array}{c}{F}_1=20\text{鈥凬}脳\cos {35}^{掳}\\ =20\text{鈥凬}脳0.819\\ =16.38\text{鈥凬}\end{array}$

Substitute 20 N for F_tot, and$35掳$for$胃$in equation (ii), and we get,

$\begin{array}{c}{F}_2=20\text{鈥凬}脳\sin {35}^{掳}\\ =20\text{鈥凬}脳\text{0.57357}\\ =\text{11.47鈥凬}\end{array}$

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

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

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

Let F₁鈥� be the force at an angle of $20掳$with the horizontal. Then, calculate its magnitude.

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

Substitute 16.38 N for F₁.

$\begin{array}{c}{F}_1\text{'}=\frac{16.38\text{鈥凬}}{\cos {20}^{掳}}\\ =\frac{16.38}{0.9397}\\ =17.43\text{鈥凬}\end{array}$

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

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

Substitute 20 N for F_total and 17.43 N for F₁鈥�.

$\begin{array}{c}{F}_2\text{'}=\sqrt{{\left(20\text{鈥凬}\right)}^2鈭�{\left(17.43\text{鈥凬}\right)}^2}\\ =9.80\text{鈥凬}\end{array}$

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