There are several examples of connected object problems in Section 6-4. Study those before examining the following problem. The example below shows the method of solving a connected object problem. We expect you to use this method in such problems.
Problem statement
Two blocks are connected with a string passing over a pulley. Block 1 is on a horizontal plane and is being pushed by a force F to the right. Block 2 is on a plane inclined at 30° with respect to the horizontal. Both planes are frictionless, and the pulley is frictionless and massless. The string likewise is massless, and it cannot stretch. Find the acceleration of the system and the tension in the connecting string. (Additional given information is provided on the picture.)
Picture of the situation (with givens)
Goals
a) Find the acceleration of the system.
b) Find the tension in the string.
Force diagrams
Special conditions
As long as the cord is taut but doesn’t stretch and the pulley is massless and frictionless, we can say, by the 3rd law, that T1 = T2. We can therefore remove the subscript and just call the tension, T. In that case, the two masses will always remain the same distance apart and will have the same acceleration, which we will denote as ax.
Net force equations
Mass 1
Fnet,1x = F + T
Fnet,1y = N1 – m1g
Mass 2
Fnet,2x = m2gsinq - T
Fnet,2y = N2 – m2gcosq
The Fnet,y equations won’t be needed. There’s no friction, so forces along one axis don’t affect those along the other.
Applying the 2nd Law and solving for the acceleration
Applying the 2nd law to the Fnet,x equations:
m1ax = F + T
m2ax = m2gsinq - T
Solving the first equation for T and substituting into the second:
T = m1ax - F
m2ax = m2gsinq - (m1ax - F)
Solving for ax:
.
Units and sign check
All quantities to the right of the equal sign are positive, so acceleration is positive as expected. The units on the right are N/kg, also expected.
Additional checks
It makes sense that F is added to m2gsinq, the component of m2g along the inclined plane. These two forces act in the same direction and both act to increase the acceleration of the system. For the following checks, let’s simplify things by setting F = 0. Then, we have:
.
Now let’s look at the angle. If q = 0°, we expect no acceleration, since everything is horizontal. The equation above does indeed give ax = 0 when sinq = 0. If q = 90°, we get:
.
Note that if the roles of the masses are reversed, that is, m1 becomes m2 and vice versa, the equation still holds. We could call this a symmetry test.
Let’s examine the equation using special values for the masses.
Case 1. m2 = 0. The formula gives ax = 0. This makes sense, as there would be no accelerating force acting on m1.
Case 2. m1 = 0. The formula gives ax = g. With no m1, there would be no tension in the cord, so this result makes sense.
Solving for the tension
We’re now ready to solve for the tension, T. We can start with Fnet,2x and solve for T.
Substitute the expression for ax and simplify.
.
Check
Note that if F = 0 and q = 90°, we get
. This is the expected result.
(See Example Problem 6-6.)
Substituting numerical values and reducing units
