Approximately what force, FM, must the extensor muscle in the upper arm exert on the lower arm to hold a 6.9 kg shot put (Fig. 9-66)? Assume the lower arm has a mass of 2.8 kg and its CG is 13.0 cm from the pivot point.
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To determine the force exerted by the extensor muscle (FM) on the lower arm, we need to consider the torque equilibrium.
The torque exerted by an object is given by the formula:
Torque = Force × Distance
In this case, the torque exerted by the shot put around the pivot point is balanced by the torque exerted by the extensor muscle on the lower arm.
The mass of the shot put (m1) is given as 6.9 kg, and the mass of the lower arm (m2) is given as 2.8 kg.
Let's assume the distance of the shot put's center of gravity (CG) from the pivot point is d1, and the distance of the lower arm's CG from the pivot point is d2. The value for d2 is given as 13.0 cm.
Since we're looking for the force exerted by the extensor muscle (FM), we can set up the following equation:
Torque exerted by the shot put = Torque exerted by the extensor muscle
(m1 × g × d1) = (m2 × g × d2) + (FM × d2)
Here, g represents the acceleration due to gravity, which is approximately 9.8 m/s^2.
We can now substitute the given values into the equation and solve for FM:
(6.9 kg × 9.8 m/s^2 × d1) = (2.8 kg × 9.8 m/s^2 × 0.13 m) + (FM × 0.13 m)
Simplifying the equation:
(67.62 N × d1) = (34.328 Nm) + (0.13 m × FM)
To solve for FM, we need to know the distance of the shot put's CG from the pivot point (d1). Unfortunately, this information is missing in the question. Therefore, we cannot provide an exact answer without that information.
To find the force exerted by the extensor muscle, we need to calculate the torque around the pivot point caused by the weight of the shot put and the lower arm.
1. Calculate the torque caused by the weight of the shot put:
Torque1 = Weight1 x Distance1
Weight1 = mass1 x gravitational acceleration
Weight1 = 6.9 kg x 9.8 m/s^2 (gravitational acceleration)
Distance1 = distance between the pivot point and the center of gravity of the shot put
Usually, the center of gravity of a spherical object like a shot put is at the center of the sphere.
As there is no information given, let's assume the shot put is a perfect sphere and its center of mass is at its geometric center.
Distance1 = R (radius of the shot put)
2. Calculate the torque caused by the weight of the lower arm:
Torque2 = Weight2 x Distance2
Weight2 = mass2 x gravitational acceleration
Weight2 = 2.8 kg x 9.8 m/s^2
Distance2 = distance between the pivot point and the center of gravity of the lower arm
Distance2 = 13.0 cm = 0.13 m (converted to meters)
3. Find the torque required from the extensor muscle to balance the torques caused by the weight of the shot put and the lower arm.
Net Torque = Torque1 + Torque2
To maintain rotational equilibrium (no angular acceleration), the net torque should be zero.
Therefore, Torque1 + Torque2 = 0
4. Solve for the force exerted by the extensor muscle:
Torque-Extensor Muscle = FM x Distance3
Distance3 = distance between the pivot point and the point where the extensor muscle acts
Unfortunately, the exact distance is not provided in the question. Let's assume it to be the same as Distance2 (0.13 m).
Now, we can solve for FM:
FM = (Torque1 + Torque2) / Distance3
Calculating the values:
Torque1 = Weight1 x Distance1
Torque1 = (6.9 kg x 9.8 m/s^2) x R
Torque2 = Weight2 x Distance2
Torque2 = (2.8 kg x 9.8 m/s^2) x 0.13 m
Net Torque = Torque1 + Torque2
0 = (6.9 kg x 9.8 m/s^2 x R) + (2.8 kg x 9.8 m/s^2 x 0.13 m)
Finally, solving for FM:
FM = Net Torque / Distance3
Substituting the values, you can solve for FM.