Problem #5 Determine the couple moment M applied to the crankshaft that will keep the piston motionless when a 400 psi pressure acts on the top of the piston. The diameter of the piston is 3 inches and the piston slides with negligible friction in the cylinder. 400 psi c 8 in. Bo M 4 in. 2.5 in A
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- Given the figure: a.) Calculate the value of angle ‘Θ’ in degrees. b.) Calculate the value of ‘h’ in cm which puts the system in equilibrium. Thank youQ2/ A belt drive consist of two V-belts in parallel, on grooved pulleys of the size. The angle of the groove is 30°. The cross – section area of each belt is 750 and u is 0.12. The density of the belt material is 1.2 Mg/m' and the maximun stress in the material is 7 MN/m?. Answer the following questions 1- The centrifugal tension in belts is (a) increases power transmitted (b) decreases power transmitted (c) have no effect on power transmitted (d) increases power transmitted up to a certain speed and then decreasesF1 F2 a F3 A d. F4 Consider the following values: - F5 a = 13 m; b = 8 m; c = 10 m; d = 5 m; F1 = 3 kN; F2 = 11 kN; F3 = 2 kN; F4 = 20 kN; F5 =2 kN; a = 30° e = 60° B = 30°
- Problem 6 Compute the reaction forces in B and D. After solving the problem, answer the following question: "Is the force in D directed toward the right or direction? Does it make sense and why?" 8.4 kN f 8.4 kN D 2.8 m 4.5 m 4.5 m leftFind the smallest distance d for which the hook will remain at rest when acted on by the force P. Neglect the weight of the hook, and assume that the vertical wall is frictionless. ANS. 250 mmQ.3) The coefficient of friction between the 100 lb block (shown in figure below) and the incline plane is 0.25 and that between the cord and cylindrical support is 0.3. Determine the range of cylinder weight W for which the system shown below will be in equilibrium. p = 0.3 100 lb p = 0.25 25° W
- Q1) Yousif pushes toward the right on an 11.0 N force on the box in the opposite direction. The magnitude of the kinetic friction force between the box and the very smooth floor is 4.50 N as the box slides toward the right. Also, the magnitude of the normal force is 98 N. For the above information, find the following? a box with a force of magnitude 31.0 N. Sarah applies 6. The free-body diagram 7. The mass of the box 8. The box's acceleration in the x-direction 9. The box's acceleration in the y-direction 10. coefficient of dynamic friction Q 2) A rope pulls the block up the ramp, as shown in the figure. Find the following: 7. Draw a free body diagram for the block. 8. Write the equation that sums the forces on the block that are parallel to the ramp. 9. Write the equation that sums the forces on the block that are perpendicular to the ramp. 10. What is the acceleration of the block perpendicular to the ramp? 11. What is the friction force? a = 4.3 m/s? T = N A = 0.45 6.0 kg 0= 30°…A man who weighs 800 N, is climbing a uniform ladder that is 5.0 m long and weighs 180 N. The bottom of the ladder rests on the ground at A and leans against a frictionless wall at B. The ladder makes an angle of ? = 53.13° (3-4-5 right triangle) with the horizontal. The man pauses one-third of the way up the ladder. Find the normal force on the base of the ladder. Find the friction force on the base of the ladder. Find the minimum coefficient of static friction needed to prevent slipping at the base. Find the magnitude of the contact force on the base of the ladder. HINT: The components of the contact force FB at the base are the static friction force and the normal force. Find the direction of the contact force on the base of the ladder. HINT: The components of the contact force FB at the base are the static friction force and the normal force.GIVEN: W= 595 N GIVEN: ANGLE A= 44 DEGREE GIVEN: ANGLE B= 14 DEGREE ANSWER: SOLVE (A) magnitude P in Newton SOLVE (B) reaction Oy in Newton SOLVE (C) reaction Ox at point O in Newton PLS SHOW COMPLETE SOLUTION AND WRITE LEGIBLY AND SHOW THE FREE BODY THE DIAGRAM. PLS SOLVE IN EQUILIBRIUM PLS SOLVE COMPLETE SOLUTION.
- i li. In. O MMecnaniCS / مطلوب Untitled Section F1 F2 a F3 B. d F4 F5 Consider the following values: - a - 25 m; b - 5 m; a - 25 m; d - 5 m; F1 = 6 KN; F2 = 7 kN; F3 = 8 KN; F4 = 9 kN; F5 = 10 KN; a = 60°, 0 = 45°, B== 30° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 15.8 kN.m b) 74 kN.m c) 88 kN.m d) 59 kN.m e) 241 kN.m 0- 47 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 218.8 kN.m b) – 117.8 kN.m c) - 131.6 kN.m d) 98.9 kN.m e) 125.4 kN.m f) - 224.2 kN.m 3] What is the resultant moment of the five forces acting on the rod about point C? a) 186.7 kN.m b) - 143.6 kN.m c) 114.6 kN.m d) - 217.9 kN.m e) 224.1 KN.m ) - 135.3 KN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 217.8 kN.m b) – 139.5 kN.m c) 111.9 kN.m d) 206.9 kN.m e) - 127.1 kN.m f) 99.1 kN.m Activate 5] What is the moment of the force F2 about point E? Go to Setting f) 321.3 kN.m a) 121.7 kN.m…A turbine is connected to a generator by means of flange coupling that has a bolt circle diameter of 20 in. The generator output is 40 Hp, 3600 rpm and 90% efficiency. If there are 10 bolts, determine the force acting on each bolt in kN.The mass center G of the 1300-kg rear-engine car is located as shown in the figure. Determine the normal force under each tire when the car is in equilibrium. State any assumptions.