URGENT HELP PLEASE - Two Blocks on an Incline?

Question

https://tycho-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys114/autumn07/homework/04d/2blocks_incline_NWT/5.gif

The figure shows two identical blocks tied together with a string which passes over a pulley at the crest of the inclined planes, one of which makes an angle q1 = 29° to the horizontal, the other makes the complementary angle q2 = 61°.
a) If there is no friction anywhere, with what acceleration do the blocks move?

a = m/s2
b) If each block has a mass m = 0.7 kg, what is the tension in the string while they are both moving?

T = N
c) Now suppose the coefficient of sliding friction between the blocks and planes is µ = 0.03. With what acceleration do the blocks move in this case?

af = m/s2

Answer 1

For each of the blocks, there are three forces operating:

W the weight of the block - down
T the tension of the rope, parallel to the ramp
N the normal force, perpendicular to the ramp.

While T and W are the same for each block, the angles of the ramps are different so N is different for each and hence the net force will be different.

But since the blocks are tied together, the magnitude (though not the directions) of the accelerations have to be the same.

If an object of mass M is on a ramp at an angle @ to the horizontal, the force of gravity (i.e. the weight of the object) can be split into two components: one parallel to the ramp and one perpendicular.

W = M x g where g is approximately 9.8 m/s^2 on Earth
Wn = W cos @ = force normal (perpendicular) to the ramp
Wp = W sin @ = force parallel to the ramp

So for each block you can compute Wp, call the values P1 and P2. P1 is pulling to the left and P2 is pulling to the right.

The two blocks are going to moving in the direction of the larger P.

For the purposes of question a, we can consider the two blocks plus rope as a single object, with net force of P2 - P1 toward the right, and a combined mass of 2 x M. But since P1 and P2 also have a factor of M in them, we can cancel M out to get the net acceleration:

net accleration = (net force)/(combined mass)
= (P2 - P1)/(2 x M)
= ((Mg sin t2) - (Mg sin t1))/(2 x M)
= (g sin t2) - (g sin t1) = g (sin t2 - sin t1)

(Note that if angle t1 equals angle t2, the ramps are symmetric and the blocks won't move - just as we would expect.)

For part b we have to look at each block in isolation. Now we have:

T - the tension pulling up (the same on both)
Pi - the force pulling down
Ri - the net force = mass x acceleration

For one block, the net force is pulling in the same direction as the P force and the tension is in the opposite direction; for the other the net force is in the same direction as the tension and opposite to the P force.

The magnitude of the acceleration is the same in both cases, as is the mass, so the same is true of Ri.

By the nature of an ideal rope and pulley, the rope tension has the same magnitude at both ends too. So we have enough equations to solve for T:

R = Ma
R = P2 - T = T - P1

For part c, we have to use the normal forces, the Wn values, we computed earlier.

The drag force is given by D = Ck x N where:
N is the normal force
Ck is the coefficient of kinetic (sliding) friction.

The drag force always operates against the direction of motion so it lowers the magnitude of the net force

So if P1 < P2, we have the new net force = P2 - P1 - D1 - D2. As before, the combined mass is 2 x M, and you can divide out M symbolically or just put it in the computation, however you prefer.