Fm Magnitude in 90° Arm Holding 60-N Weight

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To determine the magnitude of the force exerted by the biceps muscle (Fm) while holding a 60-N weight at a 90° angle, the sum of the torques around the elbow joint must equal zero. The torque created by the weight is calculated using its distance from the pivot point (30 cm), while the torque from the biceps muscle is based on its distance (3.4 cm) from the pivot. By applying the principle of equilibrium, Fm can be calculated by setting the torque from the weight equal to the torque from the biceps muscle. A free body diagram is essential for visualizing the forces and distances involved in this static scenario. The solution involves basic principles of physics related to torque and equilibrium.
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A 60-N weight is held in the hand with the forearm making a 90° angle with the upper arm as shown in the figure below (note: the arm is held steady and is not moving). The biceps muscle exerts a force Fm that is 3.4 cm from the pivot point "O" at the elbow joint. Neglecting the weight of the arm and hand, what is the magnitude of Fm if the distance from the weight to the pivot point is 30 cm? Draw a free body diagram. http://img442.imageshack.us/img442/4935/free0in.jpg

can anyone help me on this one? thanks
 
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HINT: The sum of the torques about the pivot point must add to zero.
 
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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