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A ruck is one of the main concepts and skills that every player on the rugby team needs to know how to preform properly or their team will end up losing the game. A ruck happens when a player is tackled with the ball, the player decks to the ground for support from his teammates. Then players from his team and the opposition attempt to clear each other away from the ball to gain possession and potentially turn the ball over.
When comparing a 60kg player and a 80kg player together, we can find the Kinematics, Forces and Energy variables that occur during their ruck. The 60kg player had run 4 meters in 0.6 seconds to make it to the breakdown while the 80 kg player had run 3 meters in 0.6 seconds to make it to the breakdown. During the breakdown, the 80kg player had pushed the 60kg player 1.4 meters back in 2 seconds.
Kinematics :
First we need to find the final velocity and the acceleration of the 60kg player running to the breakdown :
Δd = 4 m Δd = ½ (Vi + Vf) Δt Δd = 4 m a = Vf - Vi / Δt
Δt = 0.6 s Vf = (2 Δd / Δt) - Vi Δt = 0.6 s = 13.33 m/s / 0.6 s
Vi = 0 m/s = [2 (4 m) / 0.6 s] - 0 m/s Vi = 0 m/s = 22.22 m/s² [Right]
Vf = ? = 13.33 m/s [Right] Vf = 13.33 m/s
a = ?
Now, we need to find the same thing, but for the 80kg player :
Δd = 3 m Δd = ½ (Vi + Vf) Δt Δd = 3 m Vf² = Vi² + 2aΔd
t = 0.6 s Vf = (2 Δd / Δt) - Vi Δt = 0.6 s a = Vf² - Vi² / 2Δd
Vi = 0 m/s = [2 (3 m) / 0.6 s] - 0 Vi = 0 m/s = (10 m/s)² / 2 (3 m)
Vf = ? = 10 m/s [Left] Vf = 10 m/s = 16.67 m/s² [Left]
a = ?
Finally, we need to find the acceleration and the initial velocity of the ruck system :
Δd = 1.4 m Δd = Vf Δt - ½ aΔt² Δd = 1.4 m a = Vf - Vi / Δt
Δt = 2 s a = 2 Δd - Vf Δt / Δt² Δt = 2 s Vi = a Δt - Vf
Vf = 0 m/s = 2 (1.4 m) - 0 / (2 s)² Vf = 0 m/s = (0.7 m/s²) (2 s) - 0 m/s
a = ? = 0.7 m/s² [Left] a = 0.7 m/s² = 1.4 m/s [Left]
Vf = ?
From what we have just learned, we can see that the maximum speed of both players are 13.33 m/s and 10 m/s while running with accelerations of 22.22 m/s² and 16.67 m/s². The rucking system has a maximum speed of 1.4 m/s and an acceleration of 0.7 m/s².
Forces :
Using the accelerations we found before, we are able to find the forces that are involved in the ruck including the force of friction by using a Mu of 0.5 for rubber on plywood. First, we start with the 60kg player in the ruck :
m = 60 kg Fnet = m a mu = 0.5 Ff = (mu) (ag) (m) Fa = Ff + Fnet
a = 0.7 m/s² = (60 kg) (0.7 m/s²) ag = 9.8 m/s² = (0.5) (9.8 m/s²) (60 kg) = (42 N) + (294 N)
Fnet = ? = 42 N m = 60 kg = 294 N = 336 N
Now we can move onto the 80kg player during the ruck :
m = 80 kg Fnet = m a mu = 0.5 Ff = (mu) (ag) (m) Fa = Ff + Fnet
a = 0.7 m/s² = (80 kg) (0.7 m/s²) ag = 9.8 m/s² = (0.5) (9.8 m/s²) (80 kg) = (56 N) + (392 N)
Fnet = ? = 56 N m = 80 kg = 392 N = 448 N
Finally, we can find the forces for the entire rucking system of 140 kg :
m = 140 kg Fnet = m a mu = 0.5 Ff = (mu) (ag) (m) Fa = Ff + Fnet
a = 0.7 m/s² = (140 kg) (0.7 m/s²) ag = 9.8 m/s² = (0.5) (9.8 m/s²) (140 kg) = (98 N) + (686 N)
Fnet = ? = 98 N m = 140 kg = 686 N = 784 N
Looking at the Free Body Diagrams, we can see that the forces that are being applied by the 80 kg player are greater than those of the 60 kg player :
When comparing a 60kg player and a 80kg player together, we can find the Kinematics, Forces and Energy variables that occur during their ruck. The 60kg player had run 4 meters in 0.6 seconds to make it to the breakdown while the 80 kg player had run 3 meters in 0.6 seconds to make it to the breakdown. During the breakdown, the 80kg player had pushed the 60kg player 1.4 meters back in 2 seconds.
Kinematics :
First we need to find the final velocity and the acceleration of the 60kg player running to the breakdown :
Δd = 4 m Δd = ½ (Vi + Vf) Δt Δd = 4 m a = Vf - Vi / Δt
Δt = 0.6 s Vf = (2 Δd / Δt) - Vi Δt = 0.6 s = 13.33 m/s / 0.6 s
Vi = 0 m/s = [2 (4 m) / 0.6 s] - 0 m/s Vi = 0 m/s = 22.22 m/s² [Right]
Vf = ? = 13.33 m/s [Right] Vf = 13.33 m/s
a = ?
Now, we need to find the same thing, but for the 80kg player :
Δd = 3 m Δd = ½ (Vi + Vf) Δt Δd = 3 m Vf² = Vi² + 2aΔd
t = 0.6 s Vf = (2 Δd / Δt) - Vi Δt = 0.6 s a = Vf² - Vi² / 2Δd
Vi = 0 m/s = [2 (3 m) / 0.6 s] - 0 Vi = 0 m/s = (10 m/s)² / 2 (3 m)
Vf = ? = 10 m/s [Left] Vf = 10 m/s = 16.67 m/s² [Left]
a = ?
Finally, we need to find the acceleration and the initial velocity of the ruck system :
Δd = 1.4 m Δd = Vf Δt - ½ aΔt² Δd = 1.4 m a = Vf - Vi / Δt
Δt = 2 s a = 2 Δd - Vf Δt / Δt² Δt = 2 s Vi = a Δt - Vf
Vf = 0 m/s = 2 (1.4 m) - 0 / (2 s)² Vf = 0 m/s = (0.7 m/s²) (2 s) - 0 m/s
a = ? = 0.7 m/s² [Left] a = 0.7 m/s² = 1.4 m/s [Left]
Vf = ?
From what we have just learned, we can see that the maximum speed of both players are 13.33 m/s and 10 m/s while running with accelerations of 22.22 m/s² and 16.67 m/s². The rucking system has a maximum speed of 1.4 m/s and an acceleration of 0.7 m/s².
Forces :
Using the accelerations we found before, we are able to find the forces that are involved in the ruck including the force of friction by using a Mu of 0.5 for rubber on plywood. First, we start with the 60kg player in the ruck :
m = 60 kg Fnet = m a mu = 0.5 Ff = (mu) (ag) (m) Fa = Ff + Fnet
a = 0.7 m/s² = (60 kg) (0.7 m/s²) ag = 9.8 m/s² = (0.5) (9.8 m/s²) (60 kg) = (42 N) + (294 N)
Fnet = ? = 42 N m = 60 kg = 294 N = 336 N
Now we can move onto the 80kg player during the ruck :
m = 80 kg Fnet = m a mu = 0.5 Ff = (mu) (ag) (m) Fa = Ff + Fnet
a = 0.7 m/s² = (80 kg) (0.7 m/s²) ag = 9.8 m/s² = (0.5) (9.8 m/s²) (80 kg) = (56 N) + (392 N)
Fnet = ? = 56 N m = 80 kg = 392 N = 448 N
Finally, we can find the forces for the entire rucking system of 140 kg :
m = 140 kg Fnet = m a mu = 0.5 Ff = (mu) (ag) (m) Fa = Ff + Fnet
a = 0.7 m/s² = (140 kg) (0.7 m/s²) ag = 9.8 m/s² = (0.5) (9.8 m/s²) (140 kg) = (98 N) + (686 N)
Fnet = ? = 98 N m = 140 kg = 686 N = 784 N
Looking at the Free Body Diagrams, we can see that the forces that are being applied by the 80 kg player are greater than those of the 60 kg player :
From what we have learned here, the 80 kg player can produce forces way greater than the 60 kg player which is what is needed during a ruck to gain the advantage on your opponent and potentially turn the ball over and score.
Energy :
During the entire movement of the ruck, the amount of kinetic energy, work and power drastically decreases from the initial run of the players to the their contact with each other. To view this dramatic loss, we have to start by finding the kinetic energies of the 60 kg player throughout the movement of the ruck :
Kinetic Energy of 60 kg Player Running : Kinetic Energy of 60 kg Player in Ruck :
m = 60 kg Ek = (0.5) (m) (V)² m = 60 kg Ek = (0.5) (m) (V)²
V = 13.33 m/s = (0.5) (60 kg) (13.33 m/s)² V = 1.4 m/s = (0.5) (60 kg) (1.4 m/s)²
Ek = ? = 5330. 67 J Ek = ? = 58.8 J
Kinetic Energy of 80 kg Player Running : Kinetic Energy of 60 kg Player in Ruck :
m = 80 kg Ek = (0.5) (m) (V)² m = 80 kg Ek = (0.5) (m) (V)²
V = 10 m/s = (0.5) (80 kg) (10 m/s)² V = 1.4 m/s = (0.5) (80 kg) (1.4 m/s)²
Ek = ? = 4000 J Ek = ? = 78.4 J
Kinetic Energy of Entire Ruck System :
m = 140 kg Ek = (0.5) (m) (V)²
V = 1.4 m/s = (0.5) (140 kg) (1.4 m/s)²
Ek = ? = 98 N
By using the the kinetic energy of the ruck system and the combined kinetic energy of both players running, we can find the efficiency of the ruck system :
Ek1 = 9330.67 N Eff = Eout / Ein x 100
Ek2 = 98 N = 98 N / 9330.67 N x 100
Eff = ? = 1.05 %
With this efficiency, we can see that while the contact was being made, almost 99% of the energy generated by the two players before the contact has been lost.
Now, we can find the work that each player is doing and that the rucking system is doing :
Work of 60 kg player in ruck : Work of 80 kg player in ruck : Work of entire rucking system :
Fnet = 42 N W = F Δd Fnet = 56 N W = F Δd Fnet = 98 N W = F Δd Δd = 1.4 m = (42 N) (1.4 m) Δd = 1.4 m = (56 N) (1.4 m) Δd = 1.4 m = (98 N) (1.4 m)
W = ? = 58.8 J W = ? = 78.4 J W = ? = 137.2 J
Finally, using the work of the two players and the system, we can find out the power that is generated in all three scenarios :
Power of 60 kg player in ruck : Power of 80 kg player in ruck : Power of entire ruck system :
W = 58.8 J P = W / Δt W = 78.4 J P = W / Δt W = 137.2 J P = W / Δt
Δt = 2 s = 58.8 J / 2 s Δt = 2 s = 78.4 J / 2 s Δt = 2 s = 137.2 J / 2 s
P = ? = 29.9 W P = ? = 39.2 W P = ? = 68.6 W
The energy transformation diagram of the ruck :
Energy :
During the entire movement of the ruck, the amount of kinetic energy, work and power drastically decreases from the initial run of the players to the their contact with each other. To view this dramatic loss, we have to start by finding the kinetic energies of the 60 kg player throughout the movement of the ruck :
Kinetic Energy of 60 kg Player Running : Kinetic Energy of 60 kg Player in Ruck :
m = 60 kg Ek = (0.5) (m) (V)² m = 60 kg Ek = (0.5) (m) (V)²
V = 13.33 m/s = (0.5) (60 kg) (13.33 m/s)² V = 1.4 m/s = (0.5) (60 kg) (1.4 m/s)²
Ek = ? = 5330. 67 J Ek = ? = 58.8 J
Kinetic Energy of 80 kg Player Running : Kinetic Energy of 60 kg Player in Ruck :
m = 80 kg Ek = (0.5) (m) (V)² m = 80 kg Ek = (0.5) (m) (V)²
V = 10 m/s = (0.5) (80 kg) (10 m/s)² V = 1.4 m/s = (0.5) (80 kg) (1.4 m/s)²
Ek = ? = 4000 J Ek = ? = 78.4 J
Kinetic Energy of Entire Ruck System :
m = 140 kg Ek = (0.5) (m) (V)²
V = 1.4 m/s = (0.5) (140 kg) (1.4 m/s)²
Ek = ? = 98 N
By using the the kinetic energy of the ruck system and the combined kinetic energy of both players running, we can find the efficiency of the ruck system :
Ek1 = 9330.67 N Eff = Eout / Ein x 100
Ek2 = 98 N = 98 N / 9330.67 N x 100
Eff = ? = 1.05 %
With this efficiency, we can see that while the contact was being made, almost 99% of the energy generated by the two players before the contact has been lost.
Now, we can find the work that each player is doing and that the rucking system is doing :
Work of 60 kg player in ruck : Work of 80 kg player in ruck : Work of entire rucking system :
Fnet = 42 N W = F Δd Fnet = 56 N W = F Δd Fnet = 98 N W = F Δd Δd = 1.4 m = (42 N) (1.4 m) Δd = 1.4 m = (56 N) (1.4 m) Δd = 1.4 m = (98 N) (1.4 m)
W = ? = 58.8 J W = ? = 78.4 J W = ? = 137.2 J
Finally, using the work of the two players and the system, we can find out the power that is generated in all three scenarios :
Power of 60 kg player in ruck : Power of 80 kg player in ruck : Power of entire ruck system :
W = 58.8 J P = W / Δt W = 78.4 J P = W / Δt W = 137.2 J P = W / Δt
Δt = 2 s = 58.8 J / 2 s Δt = 2 s = 78.4 J / 2 s Δt = 2 s = 137.2 J / 2 s
P = ? = 29.9 W P = ? = 39.2 W P = ? = 68.6 W
The energy transformation diagram of the ruck :
Here, we can see the ruck that was performed to gather the information :
From everything that we have learned here, we can see that when a player is tackled, having a larger playing in the ruck will be more beneficial to the team than a smaller player because he can generate a greater force in the ruck, have more work and create more power than the smaller player allowing his team to have more opportunities to score.