Q17PE - College Physics (Urone)

Question

Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. 

(a) What is magnitude of the acceleration of the two teams? 

(b) What is the tension in the section of rope between the teams?

 

Step-by-Step Solution

Answer

 (a) The magnitude of the acceleration of the two teams is 0.106 m/s².

(b) The tension in the rope is \(12214.8\;{\rm{N}}\).

Step 1: Theory

Apply Newton’s second law of motion.

\({F_{{\rm{net}}}} = ma\)

\(\left( {{F_2} - {F_1}} \right) = \left( {{m_1} + {{\rm{m}}_2}} \right)a\)                                           …….…………. (i)

Here, F_net is the net force, n is the number of members, m₁ is the average mass of the first team member, m₂ is the average mass of the second team member, F₁ is the force exerted by the first team members, F₂ is the force exerted by the second team members, and a is the acceleration of the two teams.

Step 2: Given data

• An average mass of team 1 = 68 kg.
• An average force exerted by team 1 =1350 N.   
• the second team’s members' average mass is= 73 kg.   
• Exerted average force by second-team =1365 N.

Step 3: (a) Determine the force of friction between the losing player’s feet and the grass.

Substitute 68 kg for m₁, 73 kg for m₂, 1350 N for F₁, and 1365 N for F₂ in equation (i), and we get, 

\(\begin{array}{c}1365\;{\rm{N}} - 1350\;{\rm{N}} = \left( {68 + 73} \right)\;{\rm{kg}} \times a\\15\;{\rm{N}} = 141\;{\rm{kg}} \times a\\a = \frac{{15\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{141\;{\rm{kg}}}}\\a = 0.106\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

 

Hence, the magnitude of the acceleration of the two teams is 0.106 m/s².

Step 4: (b) Determine the tension in the section of rope between the two teams

Apply Newton’s second law of motion,

 

\(T - n{F_1} = n{m_1}a\)

 

Here, T is the tension in the rope.

 

Substitute 68 kg for m₁, 9 for n, and 1350 N for F₁ in the above expression, and we get,

 

\(\begin{array}{c}T - 9 \times 1350\;{\rm{N}} = 9 \times 68\;{\rm{kg}} \times 0.106\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T - 12150\;{\rm{N}} = 64.872\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T = \left( {64.872 + 12150} \right)\;{\rm{N}}\\T = 12214.8\;{\rm{N}}\end{array}\)

 

Hence, the tension in the rope is\(12214.8\;{\rm{N}}\).