Suggested languages for you:

Americas

Europe

Q17PE

Expert-verifiedFound in: Page 161

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/s^{2}.

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

**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} **

- 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.

Substitute 68 kg for m_{1}, 73 kg for m_{2}, 1350 N for F_{1}, and 1365 N for F_{2} 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/s^{2}.

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 m_{1}, 9 for n, and 1350 N for F_{1} 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\(12214.8\;{\rm{N}}\).

94% of StudySmarter users get better grades.

Sign up for free