標題:
工程科學應力應變題 Ex1~Ex2 ?
Acceleration due to gravity = 10 ms-2;Calculation steps should be clearly shown;Answers correct to two decimal places.1. Determine the resultant and its position of the system of forces 200N15度角, 150N100度角, 120N-120度角 and 300N-50度角.2. ABCD is a light inextensible string passing round a smooth pulley C and... 顯示更多 Acceleration due to gravity = 10 ms-2; Calculation steps should be clearly shown; Answers correct to two decimal places. 1. Determine the resultant and its position of the system of forces 200N15度角, 150N100度角, 120N-120度角 and 300N-50度角. 2. ABCD is a light inextensible string passing round a smooth pulley C and a force F (N) applied at B in a direction perpendicular to the string maintains equilibrium. Determine the magnitude of F and the angle BCD.
最佳解答:
1) resolve the forces into x, y direction x- direction 200sin15 + 300sin50 + 150cos10 + 120cos30 =533N y- direction 200cos15 + 300cos50 - 150sin10 - 120sin30 =300N resultant (533^2+300^2)^0.5 =612N bearing 90-tan^-1(300/533) =60.6 deg 2) sorry, there seems to be no any weight, how comes a force is needed to balance?
其他解答:
工程科學應力應變題 Ex1~Ex2 ?
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發問:Acceleration due to gravity = 10 ms-2;Calculation steps should be clearly shown;Answers correct to two decimal places.1. Determine the resultant and its position of the system of forces 200N15度角, 150N100度角, 120N-120度角 and 300N-50度角.2. ABCD is a light inextensible string passing round a smooth pulley C and... 顯示更多 Acceleration due to gravity = 10 ms-2; Calculation steps should be clearly shown; Answers correct to two decimal places. 1. Determine the resultant and its position of the system of forces 200N15度角, 150N100度角, 120N-120度角 and 300N-50度角. 2. ABCD is a light inextensible string passing round a smooth pulley C and a force F (N) applied at B in a direction perpendicular to the string maintains equilibrium. Determine the magnitude of F and the angle BCD.
最佳解答:
1) resolve the forces into x, y direction x- direction 200sin15 + 300sin50 + 150cos10 + 120cos30 =533N y- direction 200cos15 + 300cos50 - 150sin10 - 120sin30 =300N resultant (533^2+300^2)^0.5 =612N bearing 90-tan^-1(300/533) =60.6 deg 2) sorry, there seems to be no any weight, how comes a force is needed to balance?
其他解答:
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