Question

EXERCISE HINTS: GETTING STARTED I I'M STUCK! A person with mass 55.0 kg stands d 1.55 m away from the wall on a x = 5.80 m beam, as shown in the figure below. The mass of the beam is 40.0 kg. Find the hinge force components and the tension in the wire. (Indicate the direction with the sign of your answer.) 30° x T= N Rx N R,= Need Help? Read It Talk to a Tutor

Answer 1

Tsin30 Tcos30 mg mbg

Take moment about the hinge

$\sum M=0$

$Tsin30*x-mg*d-m_{b}g*\frac{x}{2}=0$

$Tsin30*x=mg*d+m_{b}g*\frac{x}{2}$

$Tsin30*5.80m=55kg*9.81m/s^{2}*1.55m+40kg*9.81m/s^{2}*\frac{5.80m}{2}$

$Tsin30*5.8=55*9.81*1.55+40*9.81*\frac{5.8}{2}$

$Tsin30*5.8=1974.2625$

$T=\frac{1974.2625}{sin30*5.8}$

ANSWER: ${\color{Red} T=680.78N}$

==========================

Consider all horizontal forces

$\sum F_{x}=0$

$R_{x}+Tcos30=0$

$R_{x}=-Tcos30$

$R_{x}=-680.78N*cos30$

ANSWER: ${\color{Red} R_{x}=-589.573N}$

===========================

Consider all vertical forces

$R_{y}+Tsin30-mg-m_{b}g=0$

$R_{y}=mg+m_{b}g-Tsin30$

$R_{y}=55kg*9.81m/s^{2}+40kg*9.81m/s^{2}-680.78N*sin30$

$R_{y}=55*9.81+40*9.81-680.78*sin30$

ANSWER: ${\color{Red} R_{y}=591.56N}$

=============================