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Tension formula- Tug of war

In a game of tug of war, a single rope is pulled by two teams from the opposite ends. The team that pulls the other team to its side wins. This involves a lot of tension on the rope. Let us find the formula for tension in such a scenario in this Tension formula- Tug of war article.

Tension formula- Tug of war:

If you don’t know what a tug-of-war is, then you can watch this video to understand.

In this case, there are two teams. The persons on team 1 (left-side) have weight M1 and M2 and the persons on team2(right-side) have weights M3 and M4. Let us assume that team2 is winning and is pulling team1 with a force F2 that is greater than team1’s force F1. So, the entire system is moving toward the right.

Tension always acts opposite to the force applied. So, in the above image, you can see that the direction of tension is opposite to F1 and F2. We have assumed that there is no kinetic friction. In a real case scenario, there will a frictional force involved acting opposite to the direction of motion.

Net force on the system F_net = F2 – F1 (net force will push the system towards team2 since F2>F1)

We know F = mass x acceleration => F_net = total mass x a_net a_net = F_net /total mass
a_net =(F2-F1)/(M₁+M₂+M₃+M₄)

Tension Equation for Team1:

Tension formula- Tug of war — Free body diagram of Team 1

The forces acting on Team1 are – F1(pull by the team) and + T. F1 is negative because it is opposite to the direction of the net force. Tension is positive because it acts in direction of the net force. The net force on team1 is the sum of the forces.

F_net = F1 – T

(M₁+M₂)*acceleration = F1 – T

T = F1 – [(M₁+M₂)*acceleration]

We get the formula for tension by solving the free-body diagram of team 1. Let us also solve for team2 and arrive at a different formula for tension. The two different formulas for tension can be used for verification in the case of complicated systems.

Tension Equation for Team2:

Formula for tension free-body-diagram — Free body diagram of Team 2

The forces acting on Team2 are + F2(pull by the team) and – T. F2 is positive because it is in the same direction as the net force. Tension is negative because it acts in the opposite direction of the net force. The net force on team2 is the sum of the forces.

F_net = F2 – T

(M₃+M₄)*acceleration = F2 – T

T = F2 – [(M₃+M₄)*acceleration]

The value of tension is the same in both our equations. So, we arrive at two different formulas for tension. These equations can be used to cross-check the value of tension, both values should be equal.

Numerical problem on tension in the rope of a tug-of-war:

Now, let us look at a real-life example. Suppose two teams with two persons each with mass M1 and M2 in team1 and with mass M3 and M4 in team2. Team 2 is pulling the rope with a force of 80 N and Team 1 is pulling with 70 N force. Team 2 is winning the game at this point in time. Calculate the tension in the rope.

Team 2 is winning at this point in time. So the net system is moving towards team2 with a net force of 10 N (F2-F1). The acceleration of the system can be calculated for the system using the force formula.

acceleration = F_net /total mass
acceleration =10/200= 0.05 m/s²

Free body diagram of Team1:

Let us apply the same logic. The net force acting on team1 should be equal to the summation of all forces acting on it. Let us calculate the net force for team1 first.

F_net = mass* acceleration
F_net = (40+50)*0.05 = 4.5 N

Now, a net force of 4.5 N is acting on team1. The other forces acting on team1 are the tension and pulling force of 70 N.

F_net = T – 70

T = 4.5 + 70 = 74.5 N

Free body diagram of Team2:

Applying the same logic for team2. Let us calculate the net force for team2 first.

F_net = mass* acceleration
F_net = (40+70)*0.05 = 5.5 N

The forces acting on team1 are tension and the pulling force of 80 N.

F_net = 80 – T

T = 80 – 5.5 = 74.5 N

So, the value of tension in the rope is 74.5 N. Since in both cases we get the same value of tension hence it is verified that our solution is true.