A man m 80 kg grips a hand drill m 5 kg parallel to the gr

A man (m = 80 kg) grips a hand drill (m = 5 kg) parallel to the ground, as shown in Diagram 3.11.a. If the extensor carpi radialis (ECR), that is, the wrist muscle, counteracts the weight of the drill in order to hold it parallel to the ground, what is the force muscle (F_ECR)? Knowing F_ECR, compute the vertical force of the biceps muscle (F_B) as well as the vertical reaction force (R_y) at the elbow in order to hold the forearm parallel to the ground. If this same man grips the same drill but this time with an additional 25-cm-long 0.5 kg drill bit attached (see Diagram 3.11.b), calculate the same parameters as before (F\'_ECR, F\'_B, and R\'_y) that are now necessary to keep the drill and forearm parallel to the ground.

Solution

Taking the first case.

The vertical forces to be balanced

Hence Fb = Wa + Wd for equbilirium.

Taking moment at the wrist

Wd x 12 = Fecr x 2 Therefore Fecr = 6Wd

Taking moment aboit \'O\' Fb x 6 + Fecr x 2 - (Wa x 15+ Wdx 42) = Ry x L where lL is the distance Ry acting from \'O\'.

similarly if the deill bit is attached we can find all parameters

Fb = Wa + Wd + Wb

taking moment about wrist,

Wd x 12 + 0.5 x ( 12 + L0/2) = Fecr x2

Hence Fecr

Taking moment about \'o\' we have Fb x 6 + Fecr x 2 - (Wa x 15+ Wdx 42) - Wb x ( 42 + Lo/2) = Ry x L

where L is the vertical reactive force distance from pivot \'O\'