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Home Work #1 Due Date: 11 Feb, 2010 (Turn in your assignment at the mail box of S581 outside the ME general office) The solutions must be written on single-side.

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Presentation on theme: "Home Work #1 Due Date: 11 Feb, 2010 (Turn in your assignment at the mail box of S581 outside the ME general office) The solutions must be written on single-side."— Presentation transcript:

1 Home Work #1 Due Date: 11 Feb, 2010 (Turn in your assignment at the mail box of S581 outside the ME general office) The solutions must be written on single-side A4 papers only.

2 HW 1-Problem #1 The beam is subjected to the parabolic loading. Determine an equivalent force and couple system at point A. F=2.667kN M RA =0.667kN.m

3 HW 1-Problem #2 Two couples act on the frame. If the resultant couple moment is to be zero, determine the distance d between the 500-N couple forces. D=1.663m M1=500*0.9*sin60 M2=-500*(0.9+d)*sin60 M3=-750*(1.8+d)*3/5 M4=750*[(1.8+d)*3/5+1.2*4/5] M1+M2+M3+M4=0 D=1.663m

4 HW 1-Problem #3 The five ropes in figure can each take 1500N without breaking. How heavy can W be without breaking any? 7F=W 4F=1500N W=7*1500/4=2625N F F 2F 4F

5 HW 1-Problem #4 The man in figure weighs 800N. He pulls down on the rope, raising the 250-N weight. He finds that the higher it goes, the more he must pull to raise it further. Explain this, and calculate and plot the rope tension T as a function of θ. What is the value of the tension, and the angle θ, when the man can lift it no further? Neglect the sizes and weights of the pulleys.

6 HW 1-Problem #5 Two small balls A and B have masses m and 2m, respectively. They rest on a smooth circular cylinder with a horizontal axis and with radius R. They are connected by a thread of length 2R. Find the angles θ 1 and θ 2 between the radii and the vertical line OC for equilibrium, as well as the tension in the thread and forces exerted by A and B on the cylinder. Assume that the balls are very small and that the tension is constant. FFFaFb θ1θ1 θ2θ2 Fa*sin θ 1 =F*cos θ 1 Fa*cos θ 1 +F*sin θ 1 =m Fb*sin θ 2 =F*cos θ 2 Fb*cos θ 2 +F*sin θ 2 =2m sin θ 1 = 2sin θ 2 θ 1 =84.73 θ 2 =29.86 F=mgsin θ 1 =0.9958mg N Fa=mg cos θ 1 =0.0919mg N Fb=2mg cos θ 2 =1.7345mg N mg2mg


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