In the below case, in dynamics, how do i solve this pulley problem using newtons laws?
here is the drawing
picasaweb,google,co,uk/devanlevin/Album/photo#5219794055569750146
what is the acceleration of each of the bodies?
the hanging body has a mass of 2kg
the lying body has a mass of 3kg
the cords and pulleys have no mass and the system has no friction.
i realise that the 2kg body needs to have half the acceleration of the 3kg, because of the manner in which it is connected to the pulley but how do i work out the exact acceleration??
I cannot view your picasa site with the diagram. although I have Picasa on my computer. Others may have the same problem.
To solve this pulley problem using Newton's laws, you need to apply the principles of equilibrium and the relationship between forces and acceleration.
First, let's analyze the forces acting on each body in the system. Considering the hanging body of 2kg, the only force acting on it is the force of gravity pulling it downward with a magnitude of 2kg * 9.8m/s^2 = 19.6N.
The lying body of 3kg has two forces acting on it: the force of gravity pulling it downward with a magnitude of 3kg * 9.8m/s^2 = 29.4N and the tension force from the connecting cord pulling it upward. As the cord is inextensible, the tension force is the same throughout its length.
Now, applying Newton's second law (F = ma) to each body:
For the hanging body (2kg), the net force is the tension force pulling it upward. Since the only force acting on it is the force of gravity (19.6N) directed downward, the net force is the difference between the tension force and the force of gravity:
Tension - Force of Gravity = ma
For the lying body (3kg), the net force is the difference between the force of gravity (29.4N) directed downward and the tension force pulling it upward:
Force of Gravity - Tension = ma
Since the pulleys and cords have no mass and there is no friction in the system, the tension force in the cord will be the same throughout. Therefore, we can equate the tension forces in both equations above:
Tension - 19.6N = 2kg * a
29.4N - Tension = 3kg * a
Now, we can solve these equations to find the value of the acceleration (a).
To eliminate the tension variable, we can add the two equations together:
(Tension - 19.6N) + (29.4N - Tension) = 2kg * a + 3kg * a
Simplifying the equation gives:
9.8N = 5kg * a
Dividing both sides by 5kg gives:
a = 9.8N / 5kg
Thus, the acceleration of each body in the system is approximately 1.96m/s^2.