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

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

Start with FBD's of the hanging mass, pulley and sliding mass Hanging: 5*9.81-T15*a where T1 is the tension in the wire above the hanging mass rearrange a bit T15*(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 αa/r simplify a bit T1-T20.75*a The sliding mass T212.0*a let's collect our three equations and solve for our three unknowns: T1, T2, and a T15*(9.81-a) T1-T20.75*a T212.0*a Combining the first and last into the middle: 5*9.81/17.75a a2.76 m/s T135.23 N T233.16 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

May 28, 2018

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

May 28, 2018

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.

May 28, 2018

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