Calculate Reaction Forces for Bearings in Mechanical System

What are the reaction forces (RF1x, RF1y, RF2x, and RF2y) that the bearings need to supply in a mechanical system transmitting 30 hp at 3,450 rpm?

The reaction forces that the bearings need to supply are RF1x = 127.84 lb, RF1y = 129.6 lb, RF2x = -127.84 lb, and RF2y = -129.6 lb.

Calculating Reaction Forces for Bearings

A gear is mounted between two bearings and a sheave in a mechanical system transmitting 30 hp at 3,450 rpm. The distances from the center of the left bearing to the centers of the gear, right bearing, and sheave are:

  • Sheave: 8 inches
  • Gear: 1.75 inches
  • Right Bearing: 6 inches

The first step is to calculate the torque transmitted by the belt, given by:

Torque (T) = hp * rpm * 5252 / pi

Substitute the values:

T = 30 hp * 3450 rpm * 5252 / pi = 51480 lb-in

The tight side force (Ft) and slack side force (Fs) are calculated next:

  • Ft = T / (tight/slack ratio) = 10296 lb
  • Fs = Ft / (tight/slack ratio) = 2059.2 lb

Using the equations below, we can calculate the reaction forces:

  • RF1x = Ft * dg / L
  • RF1y = Ft * ds / L
  • RF2x = Fs * dg / L
  • RF2y = Fs * ds / L

Plugging in the values, we get:

  • RF1x = 10296 lb * 1.75 in / 10 in = 127.84 lb
  • RF1y = 10296 lb * 8 in / 10 in = 129.6 lb
  • RF2x = 2059.2 lb * 1.75 in / 10 in = -127.84 lb
  • RF2y = 2059.2 lb * 8 in / 10 in = -129.6 lb

Therefore, the reaction forces that the bearings need to supply are RF1x = 127.84 lb, RF1y = 129.6 lb, RF2x = -127.84 lb, and RF2y = -129.6 lb.

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