Answers to Forces 2 wksheet for AP 1 B, C, D, B, B, C, B, A, B, C 1982B2 a. b. T1 is in internal system force and will cancel in combined equations. Using Fexternal = mtotala gives T2 m1g m2g = (m1 + m2)a, solving yields T2 = 6600 N. Now using F = ma for the load gives – – T1 m1g = m1a and T1 = 6000 N – 2000B2 a. b. = N where N = m1g cos gives f c. constant velocity means F = 0 where Fexternal = m1g sin + m2g sin 2f Mg = 0 f – – – solving for M gives M = (m1 + m2) sin (3f)/g – d. Applying Newton’s second law to block 1 gives F = m1g sin = m1a which gives a = g sin f f/m1 – –
Answers to Forces 2 wksheet for AP 1 B, C, D, B, B, C, B, A, B, C 1982B2 a. b. T1 is in internal system force and will cancel in combined equations. Using Fexternal = mtotala gives T2 – m1g – m2g = (m1 + m2)a, solving yields T2 = 6600 N. Now using F = ma for the load gives T1 – m1g = m1a and T1 = 6000 N 2000B2 a. b. f = N where N = m1g cos gives c. constant velocity means F = 0 where Fexternal = m1g sin + m2g sin – f – 2f – Mg = 0 solving for M gives M = (m1 + m2) sin – (3f)/g d. Applying Newton’s second law to block 1 gives F = m1g sin – f = m1a which gives a = g sin – f/m1
AP%20Forces%20wksht%20from%20AP%20B%20Qs%20Answers.docx