Calculating Torque of Athlete's Arm at the Gym

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The discussion revolves around calculating the torque exerted by an athlete's arm while holding a steel ball in two different positions. The torque is calculated using the formula t = rf, where 'r' is the distance from the shoulder to the point of force application and 'f' is the force due to gravity. The first part of the problem was solved correctly, yielding a torque of 30.46 N·m. For the second part, the confusion arose regarding the use of sine and cosine functions to determine the correct horizontal distance for torque calculation. Ultimately, the correct approach was confirmed, leading to a successful resolution of the problem.
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Homework Statement


An athlete at the gym holds a 2.2 kg steel ball in his hand. His arm is 74 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his shoulder if he holds his arm in each of the following ways?
1) Parallel out to his side
2) Straight but 35 M below the horizontal


Homework Equations


t=rf


The Attempt at a Solution


I don't even know how to start this one :(
 
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Jaklynn429 said:

Homework Statement


An athlete at the gym holds a 2.2 kg steel ball in his hand. His arm is 74 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his shoulder if he holds his arm in each of the following ways?
1) Parallel out to his side
2) Straight but 35 M below the horizontal

Homework Equations


t=rf

The Attempt at a Solution


I don't even know how to start this one :(

How does the distance of the ball and center of mass of the arm from his shoulder change horizontally? Isn't torque about a point the distance away through which the force acts? Hence the vertical force of gravity acts through the ball and center of mass straight down, so how far away along the horizontal in each of the 2 situations?
 
I got the first part by using
(4*.37) + (2.2*.74) = 3.108*9.8=30.46

Now I am having trouble with the second part, wouldn't I just use that answer times sin(35)? I get 17.468 and its telling me I am wrong.
 
Jaklynn429 said:
I got the first part by using
(4*.37) + (2.2*.74) = 3.108*9.8=30.46

Now I am having trouble with the second part, wouldn't I just use that answer times sin(35)? I get 17.468 and its telling me I am wrong.

Isn't Cos the horizontal distance?
 
I got the answer! thank you so much!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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