Physics Calculation PDF

### Page 1 ``` F<sub>1</sub>=-180 L F<sub>R</sub>= 600 j F<sub>2</sub>= 167 L + 199 j - 150 X F<sub>3</sub> F<sub>R</sub>= F<sub>1</sub> + F<sub>2</sub> + F<sub>3</sub> F<sub>3</sub>= F<sub>R</sub> - F<sub>1</sub> - F<sub>2</sub> F<sub>3</sub> = 13 L + 401 j + 150 K ``` A diagram with an arrow poi...

### Page 1 ``` F<sub>1</sub>=-180 L F<sub>R</sub>= 600 j F<sub>2</sub>= 167 L + 199 j - 150 X F<sub>3</sub> F<sub>R</sub>= F<sub>1</sub> + F<sub>2</sub> + F<sub>3</sub> F<sub>3</sub>= F<sub>R</sub> - F<sub>1</sub> - F<sub>2</sub> F<sub>3</sub> = 13 L + 401 j + 150 K ``` A diagram with an arrow pointing in the x, y, and z directions is shown. The arrow in the z direction has the following labels, top to bottom, along with the values: * F_z * x= 60 * 300 053^o sin 40 * 240 * x * 300 053^o * 300 sin 30^o F₂ = 300 lb * 300 cos 30^o * F₂ = 180 * 300 cos 65^o cos 40 * F_R = 600 j ### Page 2 ``` F<sub>3</sub>= 200 K F<sub>1</sub>= (1/3 * 90)+ 0 j + (1/3 * 90) X F<sub>exy</sub>= 150 cos 60 = 75 F<sub>exz</sub>= 150 sin 60 = 129.9 F<sub>ex</sub>= 75 sin 45 = 53.03 F<sub>zy</sub>= 75 cos 45 = 53.03 F<sub>2</sub>= 53.03 i + 53.03 j + 129.2 K ``` A 3-D Diagram is shown, with an arrow that points upward in the z direction, rightward in the x-direction, and to the right, behind the origin in the y-direction. The arrow in the y-direction has the following values written: * 45^o * 60 * 60^o * 2x y ### Page 3 ``` A= 0 i + 0 j + 4 K C= -3 i + 4 j + 0 K B= -1.5 i + 2 j + 2 K D = 4 i + 6 j + 0 K BD = D - B = 5.5 i + 4 j - 2 K |BD| = sqrt(5.5<sup>2</sup> + 4<sup>2</sup> + 2<sup>2</sup>) = 7.08 |BD| = 466.2 + 339 j - 169.2 K U<sub>AB</sub> = |AB| AB = 0.777 i + 0.565 j - 0.282 X ``` #### + BD = ? 600 <br> ### Page 4 A diagram shows a circle with a weight labeled Mg suspended from it. An arrow pointing downward from the weight is labeled F_g. Another arrow pointing upwards from the circle is labeled F_s. A spring is connected to the circle with the following label: **Spring constant Stiffness K** A diagram to the right depicts a particle with various forces acting on it. The following annotations appear in the diagram: * F₁ * F₂ * F₃ * W The following text appears below the diagram: ``` equilibrium of particle ∑F<sub>x</sub>= 0 ∑F<sub>y</sub>= 0 ``` ### Page 5 A diagram of a particle with various forces acting on it is shown: * 9 KN * 5 * 3 * 36.9^o * T=? * 53^o * T sin 53.1 * T cos 53.1 * x ``` equilibrium of Particle ∑F<sub>x</sub>= 0 T= 5 * 0.5531 F < ---- T cos 53.1 - 4f = 0 ∑F<sub>y</sub>=0 T = 1.33 F < < - T cos 53.1 - 4f = 0 9 - f - T sin 53.1 = 0 < T= 1.33 x 5.49 - F - 1.33 sin 53.1 = 0 = 7.2 KN 9 = 1.667 F (F = 5.4 KN) ``` ### Page 6 A diagram of particle with various forces acting on it is shown with the following notation: * 4 * 5 * FAB * FAC sin 45 * FAC cos 45 * 45 deg * A * Mg * x * TAB * KAC * FAC * 3 * B * 4 i * 45 deg * 2 * D * 300 * 5 * Z x 9.81 * 2 x 9.81 * KBD = 40 N/m * KAD = 40 N/m * FAD = 2 x 9.81 * ALAD = 2 x 9.81 / 40 * = 0.49 * KAC = 20 N/m * TAB = 30 N/m ``` ∑F<sub>x</sub> = 0 ∑F<sub>y</sub> = 0 1/3 * FAB - FAC cos 45 = 0 3 1/3 * FAB+ FAC sin 45 - 2 x 9.81 = 0 FAB = KAB (△L)AB ⇒ 14 = 30 (△L)AB FAC = 15,86 N = KAC (△L)AC △L<sub>AB</sub> = 0.464 M △L<sub>AC</sub> = 15.86 / 20 = 0.793 M ``` ### Page 7 ``` equilibrium of particle FAB 2/3 F<sub>AC</sub> sin 45 F<sub>AC</sub> cos 45 C B 4 I 45 deg 2 D 5 3 KAB = 30 N/m KAD = 40 N/m FAD = 2 x 9.81 ALAD = 2 x 9.81 / 40 = 0.49 KAC = 20 N/m TAB = 30 N/m ```