Force Couple System: Mechanics Problems & Practical Examples

AI Thread Summary
Force systems are often replaced by force and couple systems in mechanics to simplify analysis and calculations. This approach allows for the determination of equivalent forces and moments acting on a body, making it easier to solve complex problems. Practical examples include analyzing beams under load, where distributed forces can be converted into point loads and moments. Understanding this concept is crucial for effective problem-solving in engineering and physics. The replacement enhances clarity and accuracy in mechanical analysis.
chandran
Messages
137
Reaction score
1
In mechanics problems why always the force systems are replaced by force and couple systems. any practical example is available?
 
Physics news on Phys.org
chandran said:
In mechanics problems why always the force systems are replaced by force and couple systems. any practical example is available?

Perhaps you meant why force and couple systems are replaced with equivalent resulting systems?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top