Pulley problem: find equilibrium

[bebeoshea]

Pulley Problem.

m1=10kg and m2=20kg both at angle 40 degrees.Find the force, parallel to the frictionless plan.e that will allow the systme to move at constant speed.

I would know how to solve this problem if it wasn't constant speed.but it being that just through me off.All i want is a start off.thanks in advanced.

oshea

 

[radou]

Pulley Problem.

m1=10kg and m2=20kg both at angle 40 degrees.Find the force, parallel to the frictionless plan.e that will allow the systme to move at constant speed.

I would know how to solve this problem if it wasn't constant speed.but it being that just through me off.All i want is a start off.thanks in advanced.

oshea

Look up Newton's second law and see what happens to the net force acting on a system when the speed is constant.

 

[kyle8921]

123,

To find the force required to keep the system moving at a constant speed, we need to consider the forces acting on each mass. The first thing we need to do is draw a free body diagram for each mass. This will help us visualize the forces acting on each mass and their directions.

For mass m1, we have the force of gravity (mg) acting downwards, and the tension force (T1) acting upwards at an angle of 40 degrees from the horizontal. Similarly, for mass m2, we have the force of gravity (2mg) acting downwards, and the tension force (T2) acting upwards at an angle of 40 degrees from the horizontal.

Since the system is at equilibrium, the net force on each mass must be zero. This means that the vertical components of the tension forces (T1 and T2) must be equal to the force of gravity acting on each mass (mg and 2mg). This can be represented by the following equations:

T1sin40 = mg
T2sin40 = 2mg

We can solve for T1 and T2 by dividing both equations by sin40 and substituting the values for m1 and m2:

T1 = (10kg)(9.8m/s^2)/sin40 = 155.6 N
T2 = (20kg)(9.8m/s^2)/sin40 = 311.2 N

Now, to find the force parallel to the frictionless plane that will keep the system moving at a constant speed, we need to consider the horizontal forces. Since there is no friction, the only horizontal force is the component of the tension force (T1cos40 and T2cos40) acting in the direction of motion. This can be represented by the following equation:

F = T1cos40 + T2cos40

Substituting the values for T1 and T2, we get:

F = (155.6 N)cos40 + (311.2 N)cos40 = 349.4 N

Therefore, the force parallel to the frictionless plane that will keep the system moving at a constant speed is 349.4 N.