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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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?

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

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

See the step by step solution

## 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, Fnet 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$$12214.8\;{\rm{N}}$$.