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

Answer:

The  tension on the rope  is  T  =  900 N

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

        $a = \frac{F_{net}}{m_I + m_r }$

substituting  values

     $a = \frac{170}{100 + 70 }$

     $a = 1 \ m/s ^2$

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