Question

A uniform 40.0-kg scaffold of length 6.0 m is supported by two light cables, as shown below. an 80.0-kg painter stands 1.0 m from the left end of the scaffold, and his painting equipment is 1.5 m from the right end. if the tension in the left cable is twice that in the right cable, find the tensions in the cables and the mass of the equipment.

Answer 1

Here we can say that since the scaffold is horizontal so we must have to use torque balance

So here we can say about left end point we will have

$T_r(6) = 40(9.8)(3) + 80(9.8)(1) + m(9.8)(4.5)$

$T_r = 326.67 + 7.35 m$

now torque about right end

$T_L(6) = 40(9.8)(3) + m(9.8)(1.5) + 80(9.8)(5)$

$T_L = 849.33 + 2.45 m$

now we know that tension in left is twice that of right

so now we will have

$849.33 + 2.45 m = 2(326.67 + 7.35 m)$

$196 = 12.25 m$

$m = 16 kg$

so mass of the equipment is 16 kg

now tensions are

$T_L = 849.33 + 2.45(16) = 888.53 N$

$T_r = 326.67 + 7.35(16) = 444.27 N$

Answer 2

Answer:

Tensions in the left and right ends are:

$T_l=849.33+2.45(16)=888.53\;\rm{N}\\T_r=326.67+7.35(16)=444.27\;\rm{N$ and mass of painting equipment is $16\;\rm{kg$.

Explanation:

Given: A uniform $40.0\;\rm{kg$ scaffold of length $6.0\;\rm{m$ is supported by two light cables, as shown below. an $80.0\;\rm{kg$ painter stands $1.0\;\rm{m$ from the left end of the scaffold, and his painting equipment is $1.5\;\rm{m$ from the right end.

So, finding the torque on the left and right side:

$T_l=40(9.8)3+m(9.8)(1.5)+80(9.8)5\\T_l=849.33+2.45m$

and

$T_r=40(9.8)3+80(9.8)(1)+m(9.8)4.5\\T_r=326.67+7.35m$

As given that Tension in left end is twice that of right end:

$849.33+2.45m=2(326.67+7.35m)$

$196=12.25m\\m=16\;\rm{kg}$

Here, mass of the painting equipment is $16\;\rm{kg$.

Hence, tensions in the left and right ends are:

$T_l=849.33+2.45(16)=888.53\;\rm{N}\\T_r=326.67+7.35(16)=444.27\;\rm{N$

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