Question

An athlete with a mass of 70 kg is performing a barbell back squat exercise with a load of 150 kg. From the starting position (position A), the athlete stands still prior to lowering their COM in the negative vertical direction at a rate of -1.45 m/s² (position B). Once they reach the bottom position, they hold the squat for 3 seconds (e.g. 0 m/s² – position C). After the 3 second period, the athlete raises their COM in the positive vertical direction at a rate of 2.55 m/s² (position D) back to the starting position. Solve for the net force at each position to allow the athlete to achieve the respective accelerations during the barbell back squat lift.

• Starting Position (position A): ____________________________

• Lowering COM (position B): _____________________________

• Holding at bottom for 3 seconds (position C): ______________________________________

• Raising the COM to starting position (position D): ___________________________

Answer 1

The information given:

Mass of the athlete (m) - 70 kgs

Barbel Mass (M) = 150 kg

Gravitational acceleration constant - 9.8m / s 2

So to stand with this load the athlete requires the following force

Formule for force = mass X acceleration

In the position A, the athlete is not moving, as a result, only gravity is acting on the athlete

Therefore,

Force = (m + M) g

= (70 + 150 ) x 9.8

=2156 N

This the downward force acting on the athlete. So the athlete needs to produce an equal amount of force but opposite in direction to stand with the barbel. So the force generated by the athlete at position A is 2156 N

The force in position A = 2156 N -----(1)

The force required at position A is 2156 N

Position B -

Now the athlete moves downwards

therefore,

Force = (m+M)g - (m+M)a

here (n+M)g is the force due to gravity

(m+M)a is the force generated while accelerating downwards

a = 1.45m/s2

therefore force required for lowering COM (position B) =

= (70 +150) x 9.8 - (70 + 150) x 1.45

= 2156 N - 319 N

= 1837 N

Therefore the force of 1837 N is required to lower the barbel ( position B)

Now, Holding the position at point C

Here, there the athlete is not accelerating, he is stationary. So, only gravity is acting on the athlete

Therefore,

F = (m + M) x g

= (70 + 150) x 9.8

= 2156 N is the force

Therefore, a force of 2156 N is required to hold the position C

Now, raising the COM to position D,

The formula is

F = (m + M)g + (m + M)a

a = 2.55m/s2

= (70 + 150) x 9.8 + ( 70 + 150) x 2.55

= 2156 + 561

=2717 N

The force needed to move to the starting point is 2717 N ( position D)

Final answer:

1. The force required at position A is 2156 N
2. The force of 1837 N is required to lower the barbel ( position B)
3. The force of 2156 N is required to hold the position C
4. The force needed to move to the starting point is 2717 N ( position D)