http://www.webassign.net/pse/p5-15.gif
(a) Find the acceleration in Figure P5.11a.
magnitude
(b)direction
(c)Find the acceleration in Figure P5.11b.
magnitude
(d)direction
Physics help?
2015-12-10 5:31 pm
回答 (1)
2015-12-14 7:47 am
(a) a = ? for Q = 90 degrees
ax = F1/m
ay = F2/m
a = √(ax^2 + ay^2)
= √(F1^2 + F2^2) / m
(b) Horizontal Angle = ?
= atan(F2/F1)
(c) a = ? for Q = 60 degrees
ax = (F1 + F2*cosQ) / m
ay = F2*sinQ / m
a = √(ax^2 + ay^2)
= √[(F1+F2*cosQ)^2 + (F2*sinQ)^2] / m
= √[F1^2 + 2F1*F2*cosQ + (F2cosQ)^2 + (F2*sinQ)^2] / m
= √(F1^2 + F2^2 + 2F1*F2*cosQ)
= √(F1^2 + F2^2 + 2F1*F2*cos60)
= √(F1^2 + F2^2 + F1*F2)
(d) Angle = ?
= atan(F2*sinQ/(F1+F2*cosQ))
= atan(F2*sin30/(F1+F2*cos30))
= atan(0.5*F2/(F1+F2√3/2))
= atan(F2/(2F1+F2√3))
ax = F1/m
ay = F2/m
a = √(ax^2 + ay^2)
= √(F1^2 + F2^2) / m
(b) Horizontal Angle = ?
= atan(F2/F1)
(c) a = ? for Q = 60 degrees
ax = (F1 + F2*cosQ) / m
ay = F2*sinQ / m
a = √(ax^2 + ay^2)
= √[(F1+F2*cosQ)^2 + (F2*sinQ)^2] / m
= √[F1^2 + 2F1*F2*cosQ + (F2cosQ)^2 + (F2*sinQ)^2] / m
= √(F1^2 + F2^2 + 2F1*F2*cosQ)
= √(F1^2 + F2^2 + 2F1*F2*cos60)
= √(F1^2 + F2^2 + F1*F2)
(d) Angle = ?
= atan(F2*sinQ/(F1+F2*cosQ))
= atan(F2*sin30/(F1+F2*cos30))
= atan(0.5*F2/(F1+F2√3/2))
= atan(F2/(2F1+F2√3))
收錄日期: 2021-05-01 15:38:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151210093131AAYgmO9