Given:
• Length of beam = 7 m
,
• Mass of beam = 50 kg
,
• Mass of person = 60 kg
,
• Distance of the person from the left = 3.6 m
Let's find the tensions, T1 and T2.
First make a free body sketch:
Here, the net torque = 0.
To find the tension, T1, we have:
[tex]T_1*l-m_Pg*(l-l_1)-m_bg*\frac{l}{2}=0[/tex]
Where:
l = 7 m
mp = 60 kg
g is acceleration due to gravity = 9.8 m/s²
l1 = 3.6 m
mb = 50 kg
Thus, we have:
[tex]\begin{gathered} T_1*7-60*9.8*(7-3.6)-50*9.8*\frac{7}{2}=0 \\ \\ T_1*7-1999.2-1715=0 \\ \\ T_1=\frac{3714.2}{7} \\ \\ T_1=530.6\text{ N} \\ \end{gathered}[/tex]
Therefore, the tension T1 = 530.6 N.
To find the tension T2, we have:
[tex]\begin{gathered} T_2*7-60*9.8*3.6-50*9.8*3.5=0 \\ \\ T_2*7-2116.8-1715=0 \\ \\ T_2*7=2116.8+1715 \\ \\ T_2=\frac{2116.8+1715}{7} \\ T_2=547.4\text{ N} \\ \end{gathered}[/tex]
Thererfore the tension T2 = 547.5 N
• ANSWER:
T1 = 530.6 N
T2 = 547.4 N
