Question 1182783: A string attached to a body of mass 4 kg which rests on a frictionless plane that is inclined 30o with the horizontal. The string passes over a pulley at the upper end of the plane and another body of mass 3 kg is attached to the other end of the string, this body hanging vertically. (a) Find the direction and magnitude of the acceleration of the two bodies and (b) the tension in the string.
Found 2 solutions by Solver92311, ikleyn:
Answer by Solver92311(821)   (Show Source):
Answer by ikleyn(52775)  (Show Source):
You can put this solution on YOUR website! .
A string is attached to a body of mass 4 kg which rests on a frictionless plane that is inclined 30° with the horizontal.
The string passes over a pulley at the upper end of the plane, and another body of mass 3 kg is attached to the other
end of the string, this body hanging vertically.
(a) Find the direction and magnitude of the acceleration of the two bodies
(b) Find the tension in the string.
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Part (a)
Physically, the idea of the solution is VERY SIMPLE.
To answer question (a), we should compare two forces, acting at the ends of the string.
One force is the weight of the 3 kg mass hanging vertically. This force is the product
 = 3*10 = 30 newtons.
The other force is the rolling force acting on the body on the inclined plane.
This force is  =  = 2*10 = 20 newtons.
From the comparison, the weight of the 3-kg mass is greater --- HENCE what ? - - -
+---------------------------------------------------------------+
| But of course, this 3-kg mass will move down, |
| and will pull the 4-kg mass up along the incline plane. |
+---------------------------------------------------------------+
It is the ANSWER to question (a), and this part is just COMPLETED.
Part (b)
There are two ways to solve and to answer this part.
One way is EASY. It is how the beginners start solve such problems, and I will show it to you first.
The acting force (its other name is "the net force") in the system of the two connected masses
is the difference of the two forces applied at the ends
F = - = 30 N - 20 N = 10 N.
The total mass of the system is  +  = 4 kg + 3 kg = 7 kg.
According to the Newton's second law, the acceleration of masses is a =  =  = 1.429 m/s^2.
Then the tension T is T =  =  =  =  = 25.714 newtons.
Another way is longer. It is to write the system of two equations in two unknown: the tension T and the acceleration "a".
First equation is for the 3-kg mass.
Two forces act on it: the weight of 30 N directed vertically down and the tension T directed vertically up.
So, the Newton second law for this mass is
 = 30 - T, or 3a = 30 - T. (1)
Second equation is for the 4-kg mass on inclined plane. It is
 = T - , or 4a = T - 20. (2)
So, again, the system of two equations is
3a = 30 - T (1)
4a = T - 20 (2)
You can add equations (1) and (2), eliminating T. Doing this way, you find
7a = 10, a =  = 1.429 m/s^2 for the acceleration.
By subtracting equations (1) and (2) with appropriate coefficients, you can eliminate "a" and find T
4*(30- T) = 3*(T-20),
120 - 4T = 3T - 60
120 + 60 = 3T + 7T
180 = 7T
T = = 25.714 newtons.
As you see, this second way leads you to the same answer as the first way.
The problem is just solved. The ANSWER is a = = 1.429 m/s^2 for the acceleration and = 25.714 newtons for the tension.
The solution is completed.
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When I entered the University many years ago, the ability to solve such problems
(and not only solve, but TACKLE them) was a necessary condition of passing entrance exams.
I do not think that you will find many Physics sites in the Internet, where they will explain the solution to you
in such details and so perfectly, as I did it here.
Moreover, I am almost convinced, that you will find NO ONE such site in the English segment of the Internet.
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