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The figure shows a uniform, horizontal beam of length = 7 m and mass = 50 kg that ishanging by two cables as shown. If a person of mass = 60 kg stands at 3.6 m from the leftend, what are the tensions (T1 and T2) in the cables? Write only the answer to T2 incanvas (in newtons).T1T2

The Figure Shows A Uniform Horizontal Beam Of Length 7 M And Mass 50 Kg That Ishanging By Two Cables As Shown If A Person Of Mass 60 Kg Stands At 36 M From The class=

Sagot :

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

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