A block of mass m 1 = 1.50 kg and a block of mass m 2 = 6.10 kg are connected by
ID: 2241747 • Letter: A
Question
A block of mass m1 = 1.50 kg and a block of mass m2 = 6.10 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle ? = 30.0
A block of mass m1 = 1.50 kg and a block of mass m2 = 6.10 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle ? = 30.0 degree . The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. Determine the acceleration of the two blocks. Determine the tensions in the string on both sides of the pulley.Explanation / Answer
You forgot to mention that these blocks have velocity such that block 2 is descending, or that they begin from rest. I will assume this because it is unlikely that the problem is written such that block 2 is ascending and slowing down.
Forces acting on left block:
Weight (m1*g): downward
Normal force (N1): upward
Friction (F1): leftward
Tension (T1): rightward
Forces acting on right block:
Weight (m2*g): directly downward
Normal force (N2): up perpendicular to surface
Friction (F2): up parallel to surface
Tension (T2): up parallel to surface
Vertical force balance on block 1:
N1 = m1*g
Newton's 2nd law for block 1 in horizontal:
T1 - F1 = m1*a
And substitute origin of friction (F1 = mu_k*N1):
T1 - mu_k*m1*g = m1*a
Force balance perpendicular to plane on block 2:
N = m2*g*cos(theta)
Force addition parallel to plane on block 2:
m2*g*sin(theta) - T2 - F2 = m2*a
Substitute origin of friction (F2 = mu_k*N2):
m2*g*(sin(theta) - mu_k*cos(theta)) - T2 = m2*a
Acceleration is common among blocks because of an inextensible string.
Now, let's talk about the pulley. Only the tensions apply a torque to the pulley. The axle bearing reaction force will take care of any imbalance of forces, and the axle bearing reaction force is not of interest to us.
We do assume a frictionless pulley though.
Torques acting on pulley:
tau1 = T1*R (out of the page)
tau2 = T2*R (in to the page)
Rotational Newton's 2nd law:
tau2 - tau1 = I*alpha
no-slip condition determines alpha:
alpha = a/R
Thus:
R*(T2 - T1) = I*a/R
Summarize system of equations:
T1 - mu_k*m1*g = m1*a
m2*g*(sin(theta) - mu_k*cos(theta)) - T2 = m2*a
(T2 - T1) = I*a/R^2
Solve system for T1, T2, and a (algebra not displayed):
a = g*(m2*sin(theta) - m2*mu_k*cos(theta) - mu_k*m1)/(m2 + m1 + I/R^2)
T1 = m1*g*((m2 + I/R^2)*mu_k + m2*(sin(theta) - mu_k*cos(theta)) )/(m2 + m1 + I/R^2)
T2 = m2*g*((m1 + I/R^2)*(sin(theta) - mu_k*cos(theta)) + mu_k*m1)/(m2 + m1 + I/R^2)
Our pulley is treated as a uniform disc. Thus I = M*R^2/2
Substitute and simplify:
a = g*(m2*sin(theta) - m2*mu_k*cos(theta) - mu_k*m1)/(m2 + m1 + M/2)
T1 = m1*g*((m2 + M/2)*mu_k + m2*(sin(theta) - mu_k*cos(theta)) )/(m2 + m1 + M/2)
T2 = m2*g*((m1 + M/2)*(sin(theta) - mu_k*cos(theta)) + mu_k*m1)/(m2 + m1 + M/2)
Notice that it doesn't matter what the radius of the uniform disc model pulley is?
Data:
m1:=1.55 kg; m2:=5.80 kg; theta:=30 deg; mu_k:=0.36; M:=10 kg; g:=9.8 N/kg;
Results:
A) acceleration of blocks: a = 0.4235 meters/second^2
B) Tension left of pulley: T1 = 6.125 Newtons
Tension right of pulley: T2 = 8.243 Newtons
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