Explanation Given: The magnitude of the component vector force F 1 , F 1 = 80 N . The magnitude of the component vector force F 2 , F 2 = 100 N . The magnitude of the resultant force, F R = 120 N . The angle between F ′ and F R is 45 ° The angle between F ′ and y axis is 30 ° . Show the magnitude of component of forces F 1 , F 2 , and F 3 into x , y , and z axis components as in Figure (1). Express the magnitude of component force F 1 along the x direction using Figure (1). ( F 1 ) x = F 1 cos θ i Express the magnitude of component force F 1 along the y direction using Figure (1). ( F 1 ) y = 0 j Express the magnitude of component force F 1 along the z direction using Figure (1). ( F 1 ) z = F 1 ( sin θ ) k Express the magnitude of resultant force F R along the x direction using Figure (1). ( F R ) x = F ′ ( sin 30 ° ) i = F R cos 45 ° ( sin 30 ° ) i Express the magnitude of resultant force F R along the y direction using Figure (1). ( F R ) y = F ′ ( cos 30 ° ) j = F R cos 45 ° ( cos 30 ° ) j Express the magnitude of resultant force F R along the z direction using Figure (1). ( F R ) z = F R ( sin 45 ° ) k Determine the unit vector along the line of action of F 1 . u F 1 = ( F 1 ) x + ( F 1 ) y + ( F 1 ) z F 1 (I) Determine the unit resultant vector. u R = ( F R ) x + ( F R ) y + ( F R ) z F R (II) Conclusion: Substitute F 1 cos θ i for ( F 1 ) x , 0 j for ( F 1 ) y , and F 1 ( sin θ ) k for ( F 1 ) z in Equation (I). u F 1 = F 1 cos θ i + [ 0 j ] + F 1 ( sin θ ) k F 1 = cos θ i + 0 j + ( sin θ ) k Substitute 4/5 for cos θ and 3/5 for sin θ

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ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 2.6, Problem 79P

To determine

The coordinate direction angles of F1 and FR.

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Chapter 2.6, Problem 78P

Chapter 2.6, Problem 80P

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Determine the shear flow qqq for the given profile when the shear forces acting at the torsional center are Qy=30Q_y = 30Qy=30 kN and Qz=20Q_z = 20Qz=20 kN. Also, calculate qmax⁡q_{\max}qmax and τmax⁡\tau_{\max}τmax. Given:Iy=10.5×106I_y = 10.5 \times 10^6Iy=10.5×106 mm4^44,Iz=20.8×106I_z = 20.8 \times 10^6Iz=20.8×106 mm4^44,Iyz=6×106I_{yz} = 6 \times 10^6Iyz=6×106 mm4^44. Additional parameters:αy=0.5714\alpha_y = 0.5714αy=0.5714,αz=0.2885\alpha_z = 0.2885αz=0.2885,γ=1.1974\gamma = 1.1974γ=1.1974. (Check hint: τmax⁡\tau_{\max}τmax should be approximately 30 MPa.)

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Chapter 2 Solutions

Engineering Mechanics: Statics

Ch. 2.3 - Determine the magnitude of the resultant force...Ch. 2.3 - Determine the magnitude of the resultant force....Ch. 2.3 - Determine the magnitude of the resultant force and...Ch. 2.3 - Resolve the 30-lb force into components along the...Ch. 2.3 - Resolve this force into components acting along...Ch. 2.3 - along the v axis. Prob. F2-6 Ch. 2.3 - Determine the magnitude of the resultant force FR...Ch. 2.3 - If = 60 and F = 450 N, determine the magnitude of...Ch. 2.3 - If the magnitude of the resultant force is to be...Ch. 2.3 - Determine the magnitude of the resultant force FR...

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The coordinate direction angles of F 1 and F R .

The coordinate direction angles of F1 and FR.

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