- #1
azwraith69
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Homework Statement
attached below..
Homework Equations
#1:
#2:[tex]\Sigma[/tex]Fx = max [tex]\Sigma[/tex]Fy = may
#3:[tex]\Sigma[/tex]Fx = 0 [tex]\Sigma[/tex]Fy = 0
The Attempt at a Solution
#1: none.. don't know what to do yet
#2:
[tex]\Sigma[/tex]Fx = F1cos30 + F2cos45 - f = max
ax = (F1cos30 + F2cos45 - f) / m = (10cos30 + 14.1cos45 - 10) / 9 = 0.96m/s2
[tex]\Sigma[/tex]Fy = F1sin30 + F2sin45 = may
ay = (F1sin30 + F2sin45) / m = (10sin30 - 14.1sin45) / 9 = -0.55m/s2
magnitude of a = sqrt(0.962 + (-0.55)2) = 1.1m/s2
direction of a = tan-1 [tex]\frac{-0.55}{0.96}[/tex] = -29.8o or 29.8o south of east
#3:
[tex]\Sigma[/tex]Fx = FMcos33 - FVcos= 0
FVcos[tex]\theta[/tex] = FMcos33 = 60cos33 = 50.32N
[tex]\Sigma[/tex]Fy = -FMsin33 + FVsin[tex]\theta[/tex] - w= 0
FVsin[tex]\theta[/tex] = 50N + 60sin33 = 82.68N
magnitude of FV = sqrt(50.322 + 82.682) = 96.8N
direction of FV = tan-1[tex]\frac{82.68}{50.32}[/tex] = 58.67o
please help me with number 1.. i have no idea how to solve it..
the answers on the book for number 1 are: a)4.2m/s b)29.4 m/s^2 c)4310N
is there anything wrong with numbers 2 and 3??
thanks in advance
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