A force F has the components [latex]F_x=20 \text{lb}[/latex], [latex]F_y=-30 \text{lb}[/latex] , and [latex]F_z=60 \text{lb} [/latex]. Determine its magnitude F and the angles [latex]θ_x,[/latex] [latex]θ_y,[/latex] and [latex]θ_z[/latex] it forms with the coordinate axes.

You can obtain the magnitude of F from formula (2.18): [latex]F=\sqrt{F_x^2+F_y^2+F_z^2}[/latex] (2.18) [latex]\begin{aligned}F & =\sqrt{F_x^2+F_y^2+F_z^2} \\& =\sqrt{(20 ~\mathrm{lb})^2+(-30 ~\mathrm{lb})^2+(60 ~\mathrm{lb})^2} \\& =\sqrt{4900 ~\mathrm{lb}}=70 ~\mathrm{lb}\end{aligned}[/latex] Substituting the values of the components and magnitude of F into Eqs. (2.25), the direction cosines are [latex]\cos \theta_x=\frac{F_x}{F} \quad \cos \theta_y=\frac{F_y}{F} \quad \cos \theta_z=\frac{F_z}{F}[/latex] (2.25) [latex]\cos \theta_x=\frac{F_x}{F}=\frac{20 ~\mathrm{lb}}{70 ~\mathrm{lb}} \quad \cos \theta_y=\frac{F_y}{F}=\frac{-30 ~\mathrm{lb}}{70 ~\mathrm{lb}} \quad \cos \theta_z=\frac{F_z}{F}=\frac{60 ~\mathrm{lb}}{70 ~\mathrm{lb}}[/latex] Calculating each quotient and its arc cosine gives you [latex]\theta_x=73.4^{\circ} \quad \theta_y=115.4^{\circ} \quad \theta_z=31.0^{\circ}[/latex] These computations can be carried out easily with a calculator.

Question 2.CA.5: A force F has the components Fx = 20 lb, Fy = 230 lb, and Fz......

Vector Mechanics for Engineers Statics [1686415]

A force F has the components $F_x=20 \text{lb}$ , $F_y=-30 \text{lb}$ , and $F_z=60 \text{lb}$ . Determine its magnitude F and the angles $θ_x,$ $θ_y,$ and $θ_z$ it forms with the coordinate axes.

Question Data is a breakdown of the data given in the question above.

Force components: Fx = 20 lb, Fy = -30 lb, Fz = 60 lb.

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Step 1:

Calculate the magnitude of F using the given values of the components: F = sqrt((20 lb)^2 + (-30 lb)^2 + (60 lb)^2) = sqrt(400 + 900 + 3600) = sqrt(4900) = 70 lb

Step 2:

Now, let's find the direction cosines of F using equation (2.25). The direction cosine of a component is the ratio of that component to the magnitude of the force. cos θ_x = F_x / F = 20 lb / 70 lb = 2/7, cos θ_y = F_y / F = -30 lb / 70 lb = -3/7, cos θ_z = F_z / F = 60 lb / 70 lb = 6/7

Step 3:

Calculate the angle θ_x, θ_y, and θ_z by taking the inverse cosine of their respective direction cosines. θ_x = cos^(-1)(2/7) ≈ 73.4 degrees, θ_y = cos^(-1)(-3/7) ≈ 115.4 degrees, θ_z = cos^(-1)(6/7) ≈ 31.0 degrees

These calculations can be easily done using a calculator.

Final Answer

You can obtain the magnitude of F from formula (2.18):

$F=\sqrt{F_x^2+F_y^2+F_z^2}$ (2.18)

$\begin{aligned}F & =\sqrt{F_x^2+F_y^2+F_z^2} \\& =\sqrt{(20 ~\mathrm{lb})^2+(-30 ~\mathrm{lb})^2+(60 ~\mathrm{lb})^2} \\& =\sqrt{4900 ~\mathrm{lb}}=70 ~\mathrm{lb}\end{aligned}$

Substituting the values of the components and magnitude of F into Eqs. (2.25), the direction cosines are

$\cos \theta_x=\frac{F_x}{F} \quad \cos \theta_y=\frac{F_y}{F} \quad \cos \theta_z=\frac{F_z}{F}$ (2.25)

$\cos \theta_x=\frac{F_x}{F}=\frac{20 ~\mathrm{lb}}{70 ~\mathrm{lb}} \quad \cos \theta_y=\frac{F_y}{F}=\frac{-30 ~\mathrm{lb}}{70 ~\mathrm{lb}} \quad \cos \theta_z=\frac{F_z}{F}=\frac{60 ~\mathrm{lb}}{70 ~\mathrm{lb}}$

Calculating each quotient and its arc cosine gives you

$\theta_x=73.4^{\circ} \quad \theta_y=115.4^{\circ} \quad \theta_z=31.0^{\circ}$

These computations can be carried out easily with a calculator.