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 Determine the force  \(P\)  needed to hold the  \(300-N\)  weight in equilibrium. 

 

 

\( +\uparrow \sum F_ y=0 \)

\( 3 P-300=0 \)

\( \rightarrow \quad 3 P=300 \)

\(∴ P=100 N \)

 Determine the force  \(\mathbf{P}\)  needed to support the  \(50-kg\)  weight. Each pulley has a weight of  \(50 N\)  . Also, what are the cord reactions at  \(A\)  and  \(B\) ?

 

 

\( w=50 * 9.81=490.5 N \)

D

\( \uparrow \quad \sum F_{y}=0 \quad \longrightarrow T-P-P-50=0 \)

\( T=2 P+50 \)

B

\( +\uparrow \sum F_ y=0 \)

\(P+P+(2P+50)-W-50=0\)

\( 4 P=490.5 \longrightarrow P=122.6 N \)

\(∴ T_{A}=P=122.6 N \)

\( T_{B}=2 P+50=2(122 \cdot 6)+50=295.25 N \)

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