Question

Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons

Answer 1

The tension on the rope is T = 900 N

Step-by-step explanation:

From the question we are told that

The mass of the person on the left is
`m_l = 100 \ kg`

The force of the person on the left is
`F_l = 1000 \ N`

The mass of the person on the right is
`m_r = 70 \ kg`

The force of the person on the right is
`F_r = 830 \ N`

Generally the net force is mathematically represented as

`F_(Net) = F_l - F_r`

substituting values

`F_(Net) = 1000-830`

`F_(Net) = 170 \ N`

Now the acceleration net acceleration of the rope is mathematically evaluated as

`a = (F_(net))/(m_I + m_r )`

substituting values

`a = (170)/(100 + 70 )`

`a = 1 \ m/s ^2`

The force
`m_i * a`) of the person on the left that caused the rope to accelerate by a is mathematically represented as

`m_l * a = F_r -T`

Where T is the tension on the rope

substituting values

`100 * 1 = 1000 - T`

=> T = 900 N