I have this homework question and was able to set up the free body diagrams for
ID: 1279768 • Letter: I
Question
I have this homework question and was able to set up the free body diagrams for the two blocks but got stuck forming the equations in part a. Please help if you can and give explanations - I want to understand the concept for my test coming up.
A block with mass M is pushed with a force F0 at an angle theta against a frictionless wall. It is connected to another hanging block with mass m by an ideal string and pulleys.
a. Apply Newton's Law in each direction for the two blocks to find 4 equations using the symbols given above.
b. Find an expression for the acceleration of m (the hanging block).
c. What does your answer in part b predict for the acceleration if the hand stops pushing? Does this make sense?
d. What is the acceleration of m if theta = 23.5 degrees, m = 3.2 kg, M = 2.1 kg, and F0 = 4.9 N?
e. While moving with the acceleration found in part d, the 2.1 kg block suddenly encounters a patch of friction with known coefficients (Us = .3 and Uk = .12). What would be the change in its acceleration?
f. Now suppose the blocks are stopped on the friction patch and the mass of M is changed. For a constant applide force (F0 = 4.9 N), what value of M would cause the blocks to remain stationary but just be on the verge of sliding (i.e. begin and remain at rest but the slightest nudge would set them in motion)? There are a few acceptable answers.
* If you could just tell me whether or not the two blocks have the same or different accelerations for part b, that would help a lot.
Explanation / Answer
IDEAL PULLEY ASSUMPTIONS:
1) String, pulleys are massless, frictionless
2) String does not stretch or compress
3) T1 = T2 = just plain T
3) Kinematics for puleys: delta-d1 = -delta-d2; v1 = -v2; a1= -a2. Meaning, for instance, "up 10 units, down ten units"; "up 10 units/s, down 10 units/s"; "up 10 units/s^2, down 10 units/s^2".
a) for the *net* force on M in x and y directions (no friction):
Fnet_My = M*a_y1 = sin(theta)F + M*g + T
Fnet_Mx = M*a_x1 = cos(theta)F + N = 0
for the *net* force on m in x and y directions (no friction):
Fnet_my = m*ay2 = m*g + T
Fnet_mx = m*ax2 = 0
b) (Fnet_my) = m*ay2 = m*g + T
ay2 = (m*g + T)/m
c) solve topmost equation in a) for T, note ay1 = - ay2 the term involving F no longer exists. Then Plug T into equation in b):
M*ay1 = T + Mg
M*(-ay2) = T + Mg
T = -M(g+ay2)
ay2 = (m*g + T)/m
m*ay2 = m*g -M*g - M*ay2
m*ay2 + M*ay2 = g(m-M)
ay2 = g(m-M)/(m+M).
The smaller block accelerates upwards, as expected, since g is a negative number and M>m.
d) Now put the vertical component of F back in and do the same thing:
M*ay1 = Fsin(theta) + T + Mg
T = M(-ay2) - Fsin(theta) - Mg
m*ay2 = m*g + T
m*ay2 = m*g + (-M*ay2 - Fsin(theta) - Mg)
ay2 = (m*g - Fsin(theta) - Mg)/(m+M)
ay2 = [(2.1)(-9.8) - (4.9N)sin(23.5