Pulley Question .. the force exerted on the axle by the pulley.?

Question

Not quite sure how to approach this problem.. not even really sure if I know which part the axle is.. any and all help is greatly appreciated.

A 12.0kg box resting on a horizontal, frictionless surface is attached to a 5.00kg weight by a thin, light wire that passes over a frictionless pulley (View Figure). The pulley has the shape of a uniform solid disk of mass 1.50kg and diameter 0.540m.

After the system is released, find magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

btw.. the moment of inertia for a disk is I=(MR^2)/2

Figure: http://session.masteringphysics.com/problemAsset/1037926/4/YF-10-44.jpg

Thank you in advance!

oops sorry for the confusion. the question I wrote at the top contradicts with the question I pasted.. what I'm trying to find is the force exerted on the PULLEY by the axle.

Answer 1

Start with FBD's of the hanging mass, pulley and sliding mass
Hanging:

5*9.81-T1=5*a
where T1 is the tension in the wire above the hanging mass
rearrange a bit
T1=5*(9.81-a)

The pulley (note that the axle does not introduce a torque on the pulley)
0.540*(T1-T2)=
(1.50*0.540^2/2)*(a/0.540)

note also that α=a/r

simplify a bit
T1-T2=0.75*a

The sliding mass

T2=12.0*a

let's collect our three equations and solve for our three unknowns:
T1, T2, and a

T1=5*(9.81-a)
T1-T2=0.75*a
T2=12.0*a
Combining the first and last into the middle:
5*9.81/17.75=a
a=2.76 m/s
T1=35.23 N
T2=33.16 N

Okay, now lets look at the vertical and horizontal forces on the axle of the pulley
Vertical:
T1+1.5*9.81
49.95 N
So the axle reacts with an upward force of 49.95 N
Horizontal
T2
33.16 N
The axle reacts with a force to the right of 33.16 N

j

Answer 2

evaluate the three products (block a million, block 2, and the pulley) one after the different. additionally keep in mind that by way of fact the rope does not slip, the acceleration of each merchandise is equivalent, we purely must be careful on the subject of the indications. Write sum of torques approximately axis of pulley (f is the torque of the axle friction): R*T1 - R*T2 - f = I*alpha Use sum of forces for block a million (4.0 kg) to get expression for T1, being careful to evaluate sign of acceleration (a is down): T1 - m1g = m1(-a) => T1 = m1g - m1a Use sum of forces for block 2 (2.0 kg) to get expression for T2, being careful to evaluate sign of acceleration (a is up): T2 - m2g = m2a => T2 = m2g + m2a replace for T1 and T2 into sum of torques and keep in mind alpha = a/R, and that i for a disk = 0.5MR^2, being careful to evaluate sign of pulley's acceleration (counterclockwise is beneficial) R(m1g - m1a) - R(m2g + m2a) - f = (0.5MR^2)(a/R) Simplify and remedy for a: a = (Rm1g - Rm2g - f) / (0.5MR + Rm1 + Rm2) remedy for time from simplified linear kinematics equation: t = (2h/a)^0.5 a = a million.6 m/s^2 t = a million.a million s

Answer 3

Yes, the previous post is the way to do it.
This one more verbosely posts a section of the problem.

(T1 - T2) = Force to make the pulley spin
>>T = torque = Force*radius = I*alpha_angular acceleration
>>>>alpha = (tangential acceleration / radius)
>>>>alpha = a/r
>>T = F*r = I*a/r = (0.5*m*r^2)*(a/r) = 0.5*m*a/r
>>T = F*r = 0.5*m*a/r
>> F = 0.5*m*a
(T1 - T2) = 0.5*mass_disk*a
(T1 - T2) = 0.5*1.5*a = 0.75*a

Also, you could do KE = PE to check calcs,
..but again, the previous post is right.