A 1.20 kg object is subjected to three forces that give it an acceleration a Ove
ID: 1874722 • Letter: A
Question
A 1.20 kg object is subjected to three forces that give it an acceleration a Overscript right-arrow EndScripts equals left-parenthesis 1.30 m divided by s Superscript 2 Baseline right-parenthesis i Overscript EndScripts plus left-parenthesis 6.60 m divided by s Superscript 2 Baseline right-parenthesis j Overscript EndScripts. If two of the three forces are Upper F Overscript right-arrow EndScripts Subscript 1 Baseline equals left-parenthesis 21.0 N right-parenthesis i Overscript EndScripts plus left-parenthesis 15.0 N right-parenthesis j Overscript EndScripts and Upper F Overscript right-arrow EndScripts Subscript 2 Baseline equals left-parenthesis 12.0 N right-parenthesis i Overscript EndScripts plus left-parenthesis 6.50 N right-parenthesis j Overscript EndScripts, find (a) the x component and (b) the y component of the the third force.
Explanation / Answer
given
m = 1.20 kg
F1 = 21 i + 15 j N
F2 = 12 i + 6.50 j N
net acceleration a = 1.30 i + 6.60 j m/s2
now ,
net force F = m*a = 1.20*[ 1.30 i + 6.60 j ] = 1.56 i + 7.92 j N
F = F1 + F2 + F3
F3 = F -F1-F2
F3 = 1.56 i + 7.92 j - [ 21 i + 15 j] - [12 i + 6.50 j]
F3 = - 31.44 i - 13.58 j N
(a)
x - component = -31.44 N
(b)
y - component = -13.58 N
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