Are the friction forces on A and B going to be the equal or not?
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Fall 2008 -- Purdue University -- West Lafayette, IN
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15 comments:
There are some subtle things about this question that addresses your concern.
* There needs to be friction acting on at least one set of wheels (without friction, the truck cannot accelerate to the right).
* I would assume that only one set of wheels are propelling the truck.
* The statement of the problem also says that the rotational inertia of the truck wheels can be neglected. This says that the friction acting on the non-driven wheels is negligible (To see this, draw an FBD of a non-driven wheel. From Euler's equation on this FBD you can see that the friction must be zero if the rotational inertia is zero.)
In summary, assume either front wheel or rear wheel drive. Put friction on the driven wheel. Do not put friction on the non-driven wheel.
Good question.
just a couple questions:
1. We are given the radius of gyration for the empty spool, but is it the same value for the 150 turns?
2. I'm thinking the tension in the cable will not be great enough to cause the center of mass of the truck to be lifted, therefore Alpha_G will be zero?
3. Since the truck is only moving in the X direction, Acceleration_Y is zero as well?
4. Even though the Spool (point S) is attached to the truck, will there be forces X_S,Y_S which act on the central axis of the spool?
In response...
1. I assumed it was the same.
2. Alpha_G = 0
3. acceleration in y direction is 0
4. I don't think you have to consider the forces acting on the axis of the spool since it is within the system.
Do you need to find angular acceleration of the spool to find the tension by taking the moment about the axis of the spool? If so, how do you find it?
so far i have four equations. 2 sum of forces and 2 moments. One moment about the truck's center of mass and another about the spool. Know i'm just trying to find an equation to find Alpha_spool, but i haven't had any luck yet.
To sully and mrb:
I agree with all that you say except the first point. The wrapped cable forms a "thin ring" having a radius of 0.75 meters. The mass moment of inertia for a ring is given by: I = m*r^2, where m is the mass of the ring and r is the radius of the ring.
The mass of the wrapped cable is the mass/length times the circumference of the ring times 150 turns (here, m = (0.75 kg/meter)*(2*pi*0.75 meters/turn)*(150 turns).
I still don't see how to get the angular acceleration of the spool. Am I missing something?
I am confused about the I_G of the spool:
So the M_/G of the ENTIRE Spool (spool and cable) = -T(.750m) = I_G(alpha) so the I_G of the whole spool is equal to m^2K_G + m(r^2)?
Consider the following process in solving this problem:
FBDs: for drum and truck individually
NEWTON/EULER:
Spool: sum F_x = -T + O_x = m*a_truck
sum F_y = O_y - m*g = 0
sum M_O = -0.75*T = I_O*alpha
where m is total mass of spool and cable and I_O is the combined MMI of the spool and cable.
Truck: sum F_x = -O_x + f_A = M*a_truck
sum F_y = -O_y -M*g + N_A + N_B = 0
sum M_G = 1.8*N_B - 1.8*N_A + 1.05*f_A + 1.05*O_y + 1.4*O_x = 0
where M is the mass of the truck.
KINEMATICS
The critical point in the kinematics is that point E on the drum where the cable comes off the back has ZERO acceleration (the cable that is not wrapped on the drum is stationary). Therefore, you can write a kinematics equation between E and O:
a_O = a_E + alpha x r_O/E = (alpha*k)x(0.75*j) = -0.75*alpha*i
From this we see that:
alpha = -a_truck/0.75
SOLVE
Now knowing alpha, you can solve the first three N/E equations for T, O_y and O_x. From these, you can then solve the second three N/E equations for f_A, N_A and N_B.
This problem sucks.
I'm having trouble conceptualizing a point on the drum having zero accel. while the drum is moving with the truck.
I assumed the contact point C of the drum and cable has accel. only in the j dir. just as the cable leaves the drum, but we don't know if the radius of the drum is the same as the radius of the wrapped around cable.
The problem only states the radius of the spool of cable and it doesn't appear that the cable would satisfy the criteria for rigid body.
oops, I meant "a rigid body"
To dayodele:
Since the cable does not slip on the drum and since the unwound portion of the cable is at rest, the contact point of the drum with the cable is the INSTANT CENTER of the drum (zero velocity and zero x-component of acceleration). Therefore, the contact point of the drum can have, at most, a y-component of acceleration. Since the system is still at rest (albeit accelerating), the y-component of acceleration of the contact point is also zero. Therefore, a_E = 0.
Since the cable is thin, the radius of the drum and the radius of the wrap of the cable on the drum are the same.
The cable is not a rigid body. However, since it does not slip on the drum, its motion is the same as the drum around which it is wrapped.
Does this help?
I vote that we turn in the other problem and review this one in class tomorrow.
Jackie, you can't just vote for whichever problem you want to turn in and spend 20 minutes in class talking about the other...
I vote for this problem.
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