B 1200 mm 7.00 N/m 600 mm 700 mm

International Edition---engineering Mechanics: Statics, 4th Edition
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Author:Andrew Pytel And Jaan Kiusalaas
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Chapter7: Dry Friction
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Bent rod ABC is supported by a pin and bracket at A and  a cable at C to D. The weight of the bent rod is 7.00 N/m and triangular distributed load is applied from point B to C as shown

1-2: See questions in image

3. What approximates the tension force?

4. What is the magnitude of the resultant of the triangular distributed load, F_BC, and its location d?

B
1200 mm
B
WAB
FBC Wac
d
1200 mm
7.00 N/m
C
600 mm
700 mm
600 mm
700 mm
3. Which of the following vector equations CORRECTLY represents the summation of forces equation for the equilibrium of the bent rod?
A. FA= (-WABI) + (-WBCI) + (-FBCI) + Tcp<0,0.700,-0.600 >
B. FA + (-WABI) + (-WBCI) + (-FBCI) + TCD <0, 0.759,-0.651 >= 0
C. FA+ (-WABI) + (-WBCI) + (-FBCI) + TCD <0,-0.759, 0.651 >= 0
D. FA+ (-WABI) + (−WBC) + (−FBCĴ) + Tcp<0,0.700,-0.600 >= 0
4. Which of the following is the correct expression resulting to the summation of moments about x-axis?
A. WAB (0.3) + WBC (0.6) + FBC (0.6)+7(0.759)(0.6) = 0
B. WAB (0.3) + WBC (0.6) + FBC (0.6) - T(0.651)(0.6) = 0
C. WAB (0.3) + WBC (0.6) +FBc (0.6) T(0.759)(0.6) = 0
D. WAR (0.3) + WRC (0.6) + FRC (0.6) + T(0.651) (0.6)=0
Transcribed Image Text:B 1200 mm B WAB FBC Wac d 1200 mm 7.00 N/m C 600 mm 700 mm 600 mm 700 mm 3. Which of the following vector equations CORRECTLY represents the summation of forces equation for the equilibrium of the bent rod? A. FA= (-WABI) + (-WBCI) + (-FBCI) + Tcp<0,0.700,-0.600 > B. FA + (-WABI) + (-WBCI) + (-FBCI) + TCD <0, 0.759,-0.651 >= 0 C. FA+ (-WABI) + (-WBCI) + (-FBCI) + TCD <0,-0.759, 0.651 >= 0 D. FA+ (-WABI) + (−WBC) + (−FBCĴ) + Tcp<0,0.700,-0.600 >= 0 4. Which of the following is the correct expression resulting to the summation of moments about x-axis? A. WAB (0.3) + WBC (0.6) + FBC (0.6)+7(0.759)(0.6) = 0 B. WAB (0.3) + WBC (0.6) + FBC (0.6) - T(0.651)(0.6) = 0 C. WAB (0.3) + WBC (0.6) +FBc (0.6) T(0.759)(0.6) = 0 D. WAR (0.3) + WRC (0.6) + FRC (0.6) + T(0.651) (0.6)=0
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