Question

A prop shaft of a heavy-duty truck has two bearings which are positioned at 3 metres from each other. The shaft must rotate at 30 rad/s when the truck travels at the speed of 40 m/s. Such a shaft carries two masses of (MA = 10 kg) and (MB = 5 kg) acting at the extremities of the arms 0.45 m and 0.6 m long, respectively. The planes in which these masses rotate are 1.2 m and 2.4 m,

respectively, from the left end bearing supporting the shaft. The angle between the arms is 60°. If the masses are balanced by two counter-masses (named X and Y) rotating with the shaft acting at a radius of 0.3 m and placed at 0.3 m from each bearing centre:

• Draw a diagram illustrating the position of masses on the shaft.
• Calculate the dynamic force in Newtons. Tabulate the results.
• Calculate the moment acting on the shaft in Newton-metre. Tabulate the results.
• Determine the magnitude of the Y balance mass using a vector diagram.
• Determine the magnitude of the X balance mass using a vector diagram.

Answer 1

 moment acting on the shaft in Newton-metre.

$F=(f_1\times l_1 + f_2\times l_2 )\times 9.81=(10 \times 0.45 +5\times0.6)\times 9.81 =73.57 Nm$