Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported b

Two transmission belts pass over a double-sheaved pulley...

Key Concepts

Static Equilibrium

This concept is based on the principle that a body at rest must have all applied forces and moments balancing to zero. In mechanical systems, such as pulleys with belts, it is essential for determining unknown reactions at supports by setting up equations for the sums of forces and moments.

Belt Tension

Belt tension is the force transmitted through a belt in motion or at rest. It acts along the belt's length, and when applied to a pulley, it serves as a concentrated load that influences both the net force and moment acting on that system.

Moment (Torque) Analysis

Moments or torques are produced when a force is applied at a distance from the axis of rotation. In systems with pulleys of different radii, the belt tensions create distinct moments that must balance in order for the system to remain in static equilibrium.

Reaction Forces at Bearings

Reaction forces are the support forces exerted by bearings or other constraints that counteract applied loads. Analyzing these forces involves decomposing both the net forces and moments acting on the axle so that the bearings can maintain the system in a state of rest.

Pulley Systems

Pulley systems use wheels or sheaves to change the direction and distribution of forces. In compound or double-sheave configurations, variable radii allow for different turning effects from the same tension force, making the analysis of forces and moments more complex in ensuring overall equilibrium.

Transcript

00:01 Hi, we're going to solve the problem given to us today where we have two pulleys at point b and point c, along a shaft between points a and d.

00:14 We're given the parameters of the radii for the two pulleys and the tension that's on each pulley, and we're asked to solve the reaction forces at a and d.

00:26 We're also given one other piece of information is that at point d, there is no thrust, which means that in the x direction at point d, that's going to equal to zero.

00:42 So we solve this type of problem by summing up the forces and the moments caused on each restraint.

00:50 I've gone and taken the components at both of the constraint points, a and d, and you'll see we have ax, y, and z, and just d -y and d -z in the problem.

01:06 If we were to just simply write the basic equation for the summation of all the forces, we'd have the forces at a, at d, the tension on the belt at b and b prime, and the tension on the belts at c and c prime.

01:29 And again, those all have to sum to zero.

01:31 To tackle a problem like this, you would want to break it down into the vector components, and we've done that here with the second equation, where ax and the i direction, or the x direction, that's the unit vector i in the x direction, ay, j, a zk, and so on, all sum up to zero again.

01:57 If that is true, if you were to sum up just in the direction of each unit vector, all of those components likewise would have to sum to zero.

02:11 And that's what we've done here.

02:13 And right away, you can see all we really have in the eye direction is just ax.

02:19 So we've already solved for that component at the restraint a, and that's going to be zero, which makes sense because at the opposite end at point d, we were given that there is no thrust that's exerted at point d.

02:34 So to have equilibrium, static equilibrium at point a and d, they both must be zero since we were given that d was zero.

02:43 The other two directions are also written here at equations 1 and 2, and we'll use those in a little bit to go and solve for a, y, d, y, d, y, a, z, and d, z, and that'll be part of the solution that we're looking for.

03:01 The second equation we need to solve force is summation of forces, where we're taking the position or moment arm vector and using a cross product for the force vector.

03:14 And i've written down all of the combinations of all the forces that are acting on this rigid body system and the moment arms or position vectors for each one.

03:29 So if we take the first one and go back to the diagram and see that for the tension on the b side of pulling b, we want the moment arm that is drawn for the position vector that is drawn between points a and b.

03:47 And that's how that's written here.

03:51 And then we just do the same thing likewise for every other force that's acting on the system.

03:56 If you were to write out all of the vectors that are needed, both the position vectors and the force vectors, it would look like this, where for ab, the position vector, we see from the diagram that in the x direction is 50 millimeters.

04:20 I didn't put the units on those, just to unclutter the diagram.

04:24 And in the z direction, it's simply the radius of the pulley.

04:33 And that's what we've written down here, 150 millimeters in the eye direction.

04:38 We didn't put the units in just yet.

04:40 And the radius of the pulley at point b.

04:45 And i think it is self -explanatory with the directions for the tension and the reaction force down at d.

04:54 I did note or should note that the moment was taken at point a.

05:01 And i did that for a particular reason to eliminate needing to write all the components at point a in that equation.

05:09 So it eliminates one set of unknowns in the equations that we have to solve for.

05:15 So now we want to write the vector forms for each one of the moment equations that we're going to sum up.

05:23 And we'll just take each one of them at a time as they're presented here in this equation.

05:30 This first one will be ab crossed by the tension at point b on the pulley.

05:39 And i've written it here for you as 15i plus rbk cross by minus tbj.

05:50 And again, that's minus because the force is going down.

05:54 And the opposite direction of y.

05:57 Here it be...

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