- #1

Kalus

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Hi,

I wish to work out the reaction forces on the following diagram:

[PLAIN]http://img227.imageshack.us/img227/7055/imag0211b.jpg [Broken]

The circle is a gear with tangential force acting at Ft and radial force acting at Fr.

A and B are bearings that take the Ft and Fr forces.

I have found the reaction forces opposing the downward Fr in the X-Y plane, which are Ay = Fr*(La/L) and By= Fr*(Lb/L)

In the X-Z plane there is the sideways reaction forces due to Ft, which are Az= Ft*(La/L) and By= Ft*(La/L).

Now, I know that is not all that is to it, because the force Ft acts at half the diameter of the gear wheel, producing a torque.

My question is, which direction does that torque go and why? One of the above reaction forces needs to be modified to take this into account.

Many Thanks,

Kalus

I wish to work out the reaction forces on the following diagram:

[PLAIN]http://img227.imageshack.us/img227/7055/imag0211b.jpg [Broken]

The circle is a gear with tangential force acting at Ft and radial force acting at Fr.

A and B are bearings that take the Ft and Fr forces.

I have found the reaction forces opposing the downward Fr in the X-Y plane, which are Ay = Fr*(La/L) and By= Fr*(Lb/L)

In the X-Z plane there is the sideways reaction forces due to Ft, which are Az= Ft*(La/L) and By= Ft*(La/L).

Now, I know that is not all that is to it, because the force Ft acts at half the diameter of the gear wheel, producing a torque.

My question is, which direction does that torque go and why? One of the above reaction forces needs to be modified to take this into account.

Many Thanks,

Kalus

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