標題:
AL MECH
發問:
最佳解答:
(a) Let R be the normal reaction and Ff be the frictional force. Resolve forces in the vertical direction: R.cos(10) = Ff.sin(10) + 80g ---------- (1) where g is the acceleration due to gravity Take moment about the centre of gravity of the cyclist R.L.sin(20) = Ff.L.cos(20) where L is the distance from the centre of gravity to the road surface hence, Ff = R.tan(20) ---------------- (2) substitute into (1): R.cos(10) = R.tan(20).sin(10) + 80g ------------ (3) solve for R then solve for Ff using (2) (b) Resolve forces in the horizontal direction: R.sin(10) + Ff.cos(10) = 80v^2/10 ------------------- (4) where v is the speed of the cyclist solve for v using results of R and Ff from part (a) (c) Let B be the angle of [beta] thus, equation (2) becomes, 0.2R = R.tan(B) i.e. tan(B) = 0.2 B = arc-tan(0.2) Using equation (3), R.cos(10) = R.tan(B).sin(10) + 80g solve for R then using (4), R.sin(10) + 0.2R.cos(10) = 80v^2/10 solve for v
其他解答:
AL MECH
發問:
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here is the question: 圖片參考：http://imgcld.yimg.com/8/n/HA00042856/o/700912200109013873385960.jpg 更新: sorry for not posting the answer: a. R = 913N, Ff=332 N b. 7.8ms^-1 c=6.3ms^-1, 11.3degree最佳解答:
(a) Let R be the normal reaction and Ff be the frictional force. Resolve forces in the vertical direction: R.cos(10) = Ff.sin(10) + 80g ---------- (1) where g is the acceleration due to gravity Take moment about the centre of gravity of the cyclist R.L.sin(20) = Ff.L.cos(20) where L is the distance from the centre of gravity to the road surface hence, Ff = R.tan(20) ---------------- (2) substitute into (1): R.cos(10) = R.tan(20).sin(10) + 80g ------------ (3) solve for R then solve for Ff using (2) (b) Resolve forces in the horizontal direction: R.sin(10) + Ff.cos(10) = 80v^2/10 ------------------- (4) where v is the speed of the cyclist solve for v using results of R and Ff from part (a) (c) Let B be the angle of [beta] thus, equation (2) becomes, 0.2R = R.tan(B) i.e. tan(B) = 0.2 B = arc-tan(0.2) Using equation (3), R.cos(10) = R.tan(B).sin(10) + 80g solve for R then using (4), R.sin(10) + 0.2R.cos(10) = 80v^2/10 solve for v
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