The first thing to do in solving problems would be to draw a force diagram; they are generally quite helpful in solving pulley problems ( and for that matter, most force-related problems).
In the following example:
A 50 kg woman dangles a 50 kg mass from the end of a rope. If she stands on a frictionless surface and hangs the mass over a cliff with a pulley, the tension in the rope will be .. ?
Start by drawing your force diagram with a "person" holding a rope which is looped over a pulley and attached to a mass. Now, with regard to tension, an important fact to know is that the tension force always faces the direction away from the object to which it is attached. In this case, a tension force would thus be directed upwards away from the 50 kg mass; and additionally, another tension force (equal in both magnitude, and direction...) would point upwards from the woman holding the other end of the rope. Finally, the weight of the mass on the end of the rope must be included: recall: W=mg; thus, the downward force on the mass is (50)(10) = 500 N.
OK, now, to set up the necessary equation to relate the forces. Since Net force = (m)(a), and in this case, the object is not moving, the net force must be equal to zero, since acceleration is zero.
Finally, select your positive and negative directions: Up= positive, down = negative. Both tension forces are directed upwards, while the weight of the object is directed downwards. Thus:
F=ma=0
2T-mg=0
2T=mg
T=mg/2
T=(50)(10) /2
T=250 N
Other types of pulley problems which ask you to determine the acceleration of the block or something of the sort do not actually require you to even consider the tension forces. For example, consider a single pulley with a rope looped around it. To one end, a mass of 10 kg is attached, and to the other, a mass of 5 kg is attached. What will be the acceleration of the blocks? The way that I saw such problems solvent in TPR, they set up multiple equations which are then put together into one. However, this method is somewhat tedious, and further, unneccesary. Instead, consider the following:
All you care about is the NET FORCE driving the system. In such a case, the tension in the rope would be considered INTERNAL FORCES of the system, and need not be considered to determine the acceleration. All that needs to be considered is the net driving force on the system, which in this case, results from the weights of the 2 masses.
Draw your force diagram so that the 10 kg mass is on the right side of the pulley, and the 5 kg mass on the left. Select clockwise to be positive, counterclockwise to be negative. Now, F= ma needs to be rewritten as:
F (driving) = mass (system) X acceleration (system)
Since the weight of the 10 kg mass would drive the system in the clockwise direction, W (10kg mass) is positive, while W (5kg mass) would be negative. Thus:
W (10kg mass) -W (5kg mass) = M (system) x acceleration
(10kg)(10m/s2)- (5)(10) = (5+10) X acceleration
NOtice that the mass of the entire system is simply the sum of the 2 masses....
Finally:
100-50 =15a
50/15=a= 3.3 m/s2.
Notice that this is far easier than setting up two equations relationg to the tension forces, then making the 2 equations equal to each other, and finally solving for acceleration. This sort of analysis, considering the net driving force on the system is useful in pulley problems such as the above 2, and in more "complicated" situations where one of the blocks attached to the rope on the pulley is placed on an inclined plane. IN that case, simply consider the weight in the "horizontal" direction as the force pulling the system in that direction.
Hope this helps!
) but i think that the way you stated it with "to the pulley" would probably be more appropriate and easier to get.
