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Free body diagrams of forces, forces expressed by their components and Newton's laws are used to solve these problems.Problems involving forces of friction and tension of strings and ropes are also included.- [Instructor] Let's solve some more of these systems problems.
a) Find the magnitude of the tension in each cord in terms of α1, α2 and m so that the system is at rest.
b) Find numerical values to the three tensions found above for α1 = 45 , α2 = 30 and m = 1 Kg.
And that's trying to prevent the system from moving. And the total mass is just five plus three, is gonna be eight kilograms, and I get the acceleration of my system.
This five kilogram mass is accelerating downward, and this force is in the opposite direction of motion. They're like, I don't understand, they're both pointing down. They would when we're using Newton's second law the way we usually use it, but when we're using this trick, what we're concerned with are forces in the direction of motion, this is an easy way to figure it out, forces in the direction of motion we're gonna call positive. Because all the motion in the system is this way, we'd find that way's positive, but this force of gravity on the three kilogram mass is the opposite direction. It's preventing the system from accelerating as fast as it would have. So if I just add this up, I get 2.45 meters per second squared.
Neglect the friction and the mass of the pulley and the string. The table of contents will list only tasks having one of the required ranks in corresponding rankings and at least one of the required tags (overall).
If you wish to filter only according to some rankings or tags, leave the other groups empty.Another force that tries to prevent it from moving is the force of gravity on the three kilogram mass.Or, one force that tries to prevent the system from moving would be this force of gravity. That's three kilograms times 9.8 meters per second squared.The hard way is to solve Newton's second law for each box individually, and then combine them, and you get two equations with two unknowns, you try your best to solve the algebra without losing any sins, but let's be honest, it usually goes wrong.So, the easy way to do this, the way to get the magnitude of the acceleration of the objects in your system, that is to say, if I wanna know the magnitude at which this five kilogram box accelerates, or that this three kilogram box accelerates, all I need to do is take the net external force that tries to make my system go, and then I divide by my total mass of my system.And then I figure out, are there any other forces making this system go? You might say, well what about this tension over here?Isn't the tension on this three kilogram mass? Not really, because that's an internal force exerted between the objects in our system and internal forces are always opposed by another internal force.So one external force would just be the force of gravity on this five kilogram mass.So I'm gonna have a force of gravity this way, and that force of gravity is just going to be equal to five kilograms times 9.8 meters per second squared, because that's how we find the force of gravity. Well, this five kilogram is gonna be the one that's pulling downward, so if the question is, I hold these masses and I let go, what's the acceleration? The table of contents will list only tasks having one of the required ranks in corresponding rankings and at least one of the required tags (overall).Several problems with solutions and detailed explanations on systems with strings, pulleys and inclined planes are presented.