A car of mass Mc is connected to mass m2 by a string. The string passes over a solid cylindrical pulley, which has a frictionless bearing, of radius R and mass M. when the system is released from rest the string doesn't slip, the car moves down the incline, and m2 moves upward through a distance h'( picture shows the box hanging a little distance off the pulley and h' starts from the middle of the box from the almost bottom of the triangle ramp where the box was hanging to the middle of the box where it actually is after moving)
simple picture below
---- o------
car/| box on right
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a) draw a free body diagram for Mc, m2, M
I drew the car Mc with the normal force, friction force and tension force, force of gravity Fmg and their components x and y.
b) Use the dynamical equations of motion to find the components of the net force aceting on and net torque acting on M.
WHAT IS A DYNAMIC EQUATION??? I looked it up online and got nowhere..can I just say that the cars are not moving or does the dynamic equation involve the movement??
How would I find the net torque acting on the pulley M?
c) Use the results of b) to derive a expression for the translational aceleration for the masses in terms of Mc, m2, M, h' and the coefficient of friction
well I don't get b) but how would I do this part from what I get in b?
Please help me..
****O
*C/*|B-
*/**|*| h'
/___|*-
well I tried to draw
a better picture...
c= car
B= box
h'= height traveled by box (lookes exactly like that with the line in the middle of the box and the line almost at the bottom with a see through box (where box was at bottom) but line for h' goes and is at the middle of that one)
O= pulley
that looks horible after entering that...just eliminate the | line in the middle of the triangle ramp and imagine that right next to the h' to make a I.
****O
*C/*|B-
*/***|*| h'
/___|*-
a) To draw the free body diagrams for each object, consider the following:
1. For Mc (the car):
- There is a tension force, T, acting upward along the incline.
- There is a normal force, N, perpendicular to the incline and pointing toward the car's surface.
- There is a force of gravity, Fmg (mass of the car multiplied by acceleration due to gravity), acting downward.
- There is a friction force, Ff, opposing the motion and acting parallel to the incline.
2. For m2 (the box):
- There is a tension force, T, acting downward.
- There is a force of gravity, Fmg (mass of m2 multiplied by acceleration due to gravity), acting upward.
3. For M (the pulley):
- There are tension forces, T1 and T2, acting on opposite sides of the pulley and tangential to the pulley's surface.
- There is a force of gravity, Fmg (mass of M multiplied by acceleration due to gravity), acting downward.
b) A dynamic equation refers to the equation of motion for an object. To find the net force acting on an object, you need to consider all the individual forces acting on it and apply Newton's second law, which states that the net force equals the mass of the object multiplied by its acceleration.
For Mc (the car), the net force can be represented as:
net force = Ff - Tsin(theta) - mCgsin(theta),
where Ff is the friction force, T is the tension force, theta is the angle of the incline, and g is the acceleration due to gravity.
For M (the pulley), the net torque can be represented as:
net torque = T1 * R - T2 * R,
where T1 and T2 are the tension forces acting on opposite sides of the pulley, and R is the radius of the pulley.
c) To derive an expression for the translational acceleration in terms of the given variables, you need to relate the net force on Mc and m2 to their masses and the coefficients of friction.
Start by equating the net force on Mc to Mc * a (acceleration):
Mc * a = Tsin(theta) - Ff.
Next, equate the net force on m2 to m2 * a:
m2 * a = T - m2 * g.
From these two equations, you can solve for the acceleration, a, in terms of the given variables Mc, m2, the coefficient of friction, theta, and g.