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- A Given: . 10 A L1 L1 = 3.5m L2 = 5.5m с H = 4.25 m . FC = 25 KN D Fc. L2 Member AC- (magnitude of force) Match the reaction at the support (FAX, FAy, FB), and the forces in each member (AC,AD, CD, BD, BC) Member BC (magnitude of force) Support FAy (include direction, up positive, down negative) Support FB (include direction, up positive, down negative) Support FAX (include direction, left negative, right positive) Member BD (magnitude of force) Member AD (magnitude of force) B Member DC (magnitude of force) 1. -25 2.-19.3 3. -15.3 4. 11.8 5. 0.00 6. 11.8 7. 15.3 8. 19.3 H 9. 20.03 kg 2 kg 7m A 4m4 15m 1m 5 m Find the magnitude of force (F) 4m 1.4 m F = ? 15 N 10 N 50 N Find the length (x) -3m- Pivot Find the magnitude of force (F₂) 15 N xm 8m. 2N 3N2. Two systems of forces are shown in the figure below. Are they equivalent? Show your entire cakulations. EE 100N 200 N 1.5m 1.5m 100 N 500 N.m 1.5m 1.5m 300 N.M O200N System 1 System 2
- A simply supported beam is subjected to two forces P and Q. Value of force P is 20 N and value of force Q is 40 N.If value of support force Ra is 30 N , then value of support force Rb is calculated as : O a . 70 N O b.30N O C60N O d . SONThe horizontal component of Force F1 (unit is in N), H1= The vertical component of Force F1 (unit is in N), V1 = The horizontal component of Force F2 (unit is in N), H2= The vertical component of Force F2 (unit is in N), V2 = The horizontal component of Force F3 (unit is in N), H3= The vertical component of Force F3 (unit is in N), V3 = The vertical component of Force F4 (unit is in N), V4 = The horizontal component of Force F5 (unit is in N), H5= The vertical component of Force F5 (unit is in N), V5 = The horizontal component of Force F6 (unit is in N), H6= The vertical component of Force F6 (unit is in N), V6 = The sum of horizontal force component for the given system (unit is in N),begin inline style sum for H of end style = The sum of vertical force component for the given system (unit is in N),begin inline style sum for V of end style = The resultant of the given force system is (unit in N), R = The direction of Resultant Force is = Short Answer question4. A meter stick has a pivot that is located at (35 cm). On this meter stick we place 200 g @ 20 cm, 200 g @ 25 cm, 350 g @70 cm and 100 g @80 cm. The meter stick has a mass of 150 g. a. How much mass needs to be placed at 55 cm in order to balance the meter stick. b. Where will you put 500 g in order to balance the meter stick?
- s.google.com/form: Test user sign up |T. Secure Edge Comp. Se Test user sign up O 0.45 MPa O 0.45 kPa 2- the unit of line load is * kN2/mm pound kN/mm2 3-On a body the effect of a force on it is depe O Position or line of action Magnitude All of these 4- The process to determination of the resultant ofWhich of the following statement is correct? Select one: O a. point functions are inexact Differential. O b. Point functions are not the properties of the system. O c. The Point Function depends on the path followed during a process as well as the end states. O d. Examples for points functions are temperature, pressure, density.1.14 In the hoist drive system shown in Fig.22 , the mass M is considered being moved upwards with negligible frictional torque. Show that the load torque and the equivalent moment of inertia are given by dwm Tm = Je dt W2 М.r Wm Je = Im + J1 + ( [J2 + M.r²] Wm Wm Om, Tm Ji Motor J2 2r Jm 02 gear Loss-free Fig.22.
- a)Consider the rope attached to the top of the scale. The internal force within the rope at 2.37 in above the scale has a magnitude of _____ lb. b)Consider the rope attached to the top of the scale. This rope also attaches to the ceiling. The reaction force at this ceiling/rope interface has a magnitude of _____ lb. c)Consider the rope attached to the former student. The internal force within the rope at 16.25 in above the student has a magnitude of _____ lb.Test 3/ Course 1/ Mechanics / ko مطلوب Untitled Section F1 F2 a F3 a F4 D F5 Consider the following values: - a - 25 m; b - 5 m; a- 25 m; d -5 m; F1 - 7 kN; F2 - 8 KN; F3 - 9 kN; F4 - 10. kN; P5 - 11 kN; a= 30°, e= 60° B= 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 158 kN.m b) 127 kN.m 9146 KN.m d)- 103 KN.m e) 241 kN.m )- 171 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 98.8 kN.m b)- 291.7 KN.m ) 311.6 kN.m d)- 264.6 kN.m e) 354.2 kN.m 0-242.5 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 86.7 kN.m b) - 141.1 kN.m e) 146 KN.m d) - 272.1 kN.m e) 174.1 KN.m f) - 315.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) - 117.8 kN.m b) 82.7 KN.m ) 121.9 EN.m d)- 259.9 kN.m e) 127.1 KN.m H 300.1 kN.m 5] What is the moment of the force F2 about point E? Activate a) 151.7 kN.m b) 118.1 kN.am c) 214.6 kN.m d)…PHYS X PHYS X 印 PHYS X PHYS X POTPHYS X PHYS X E PHYS X E PHYS top/semester2/physics%20for%20engineers/PHYS220_CH15_Lecture%20Notes_Problems%2015 19,15.29 S (D Page view A Read aloud V Draw Problem-15-19: page-475 A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed, and acceleration. (b) the speed and the acceleration when the object is 6.00 em from the equilibrium position, and (c) the time interval required for the object to move from.r50 to r5 8.O0 cm. Solution: