(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

attached below..

2. Relevant equations

#1:

#2:[tex]\Sigma[/tex]F_{x}= ma_{x}[tex]\Sigma[/tex]F_{y}= ma_{y}

#3:[tex]\Sigma[/tex]F_{x}= 0 [tex]\Sigma[/tex]F_{y}= 0

3. The attempt at a solution

#1: none.. don't know what to do yet

#2:

[tex]\Sigma[/tex]F_{x}= F_{1}cos30 + F_{2}cos45 - f = ma_{x}

a_{x}= (F_{1}cos30 + F_{2}cos45 - f) / m = (10cos30 + 14.1cos45 - 10) / 9 = 0.96m/s^{2}

[tex]\Sigma[/tex]F_{y}= F_{1}sin30 + F_{2}sin45 = ma_{y}

a_{y}= (F_{1}sin30 + F_{2}sin45) / m = (10sin30 - 14.1sin45) / 9 = -0.55m/s^{2}

magnitude of a = sqrt(0.96^{2}+ (-0.55)^{2}) =1.1m/s^{2}

direction of a = tan^{-1}[tex]\frac{-0.55}{0.96}[/tex] = -29.8^{o}or29.8^{o}south of east

#3:

[tex]\Sigma[/tex]F_{x}= F_{M}cos33 - F_{V}cos= 0

F_{V}cos[tex]\theta[/tex] = F_{M}cos33 = 60cos33 = 50.32N

[tex]\Sigma[/tex]F_{y}= -F_{M}sin33 + F_{V}sin[tex]\theta[/tex] - w= 0

F_{V}sin[tex]\theta[/tex] = 50N + 60sin33 = 82.68N

magnitude of F_{V}= sqrt(50.32^{2}+ 82.68^{2}) =96.8N

direction of F_{V}= tan^{-1}[tex]\frac{82.68}{50.32}[/tex] =58.67^{o}

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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# Homework Help: Basic mechanics problems

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