Question

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
Find the tensions in the two vertical ropes supporting the scaffold.

Answer 1

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

Answer 2

The tensions in the two vertical ropes supporting the scaffold are 710.5 Newton and 465.5 Newton respectively.

Given the following data:

• Mass of person = 50 kg

• Mass of scaffold = 70 kg

• Length of scaffold, L = 6 meter

• Distance from scaffold = 1.5 meter

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$

To find the tensions in the two vertical ropes supporting the scaffold:

First of all, we would determine the resultant vertical forces acting on the system.

$T_p + T_s = m_pg + m_sg$

Where:

• $T_p$ is the tension acting on the person.

• $T_s$ is the tension acting on the scaffold.

• g is the acceleration due to gravity.

• $m_p$ is the mass of the person.

• $m_s$ is the mass of the scaffold.

Substituting the given parameters into the formula, we have;

$T_p + T_s = 50\times9.8 + 70\times9.8\\\\T_p + T_s = 490 + 686\\\\T_p + T_s = 1176\; Newton$.....equation 1.

The distance of the person from the other end of the scaffold is:

$l_2 = 6 - 1.5\\\\l_2 = 4.5\;meter$

The distance of the scaffold from the center of mass of the two vertical ropes:

$l_c = \frac{1}{2} \times distance\\\\l_c = \frac{1}{2} \times 6\\\\l_c = 3\;meter$

Since the system is at equilibrium, its moment would be equal to zero (0).

Taking moment about the two vertical ropes with respect to the tension in the scaffold ($T_s$):

$LT_p - m_pgl_2 - m_sgl_c = 0\\\\6T_p - 50\times9.8(4.5) - 70\times9.8(3) =0\\\\6T_p - 2205 - 2058 =0\\\\6T_p - 4263 = 0\\\\6T_p = 4263\\\\T_p = \frac{4263}{6}\\\\T_p = 710.5 \; Newton$

From equation 1:

$T_p + T_s = 1176\\\\710.5 + T_s = 1176\\\\T_s = 1176 - 710.5\\\\T_s = 465.5\;Netwon$

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