Force on the biceps when throwing an object in the air

Question

A waiter holds two plates of food in one hand. His forearmhas a mass of 2.2 kg and the centre of mass of his forearm is located 13 cm from his elbow joint. The centre of mass of the two plates is located 37 cm from his elbow joint and the total mass of the two plates is 1.1 kg. His bicep is attached to the bones of his forearm

3.5 cm from his elbow joint. After an argument with the chef, the waiter proceeds to throw the plates of food he is carrying straight up into the air. If the plates are accelerating at a rate of 2.5m/s^2, and the center of mass of his arm is accelerating upwards at 0.9m/s^2, what is the force being applied by the waiter’s bicep? (Assume that the waiter’s forearmalso accelerates upwards at the same rate).

Answer 1

I find the statements about the forearm's acceleration confusing.

First of all, the acceleration of the plates and the forearm should be in the

same ration as their distance from the elbow.

Therefore the acceleration of the forearm should be derivable from the

other quantities, and indeed we would get

2.5 × 13 / 37 = 0.88

which is approximately the given 0.9 m/s².

Secondly, I do not understand the the last sentence in parentheses;

so I will ignore it.

Let

m = 2.2 kg be the mass of the forearm,

d = 13 cm = 0.13 m be the distance of the forearm's center of gravity from the elbow,

M = 1.1 kg be the weight of the plates,

D = 37 cm = 0.37 m be the distance of the plates from the elbow,

b = 3.5 cm = 0.035 m be how far from the elbow the bicep is attached,

A = 2.5 m/s² be the acceleration of the plates,

a = 0.9 m/s² be the acceleration of the forearm,

g = 9.81 m/s² be gravitational acceleration,

F (to be computed) be the force of the biceps.

I am going to assume that the above data applies to the point in time when

the forearm is horizontal.

The force F generates torque on the forearm of magnitude

τ = Fb

This torque needs to be big enough to support the weight of the forearm and

the plates, and also to cause the acceleration.

Therefore

τ = mgd + MgD + mad + MAD

where

mg is the weight of the forearm, and mgd is the torque of gravity on the forearm,

Mg is the weight of the plates, and MgD is the torque of gravity on the plates,

ma is the force needed to accelerate the forearm, and mad is the torque needed to accelerate the forearm,

MA is the force needed to accelerate the plates, and MAD is the torque needed to accelerate the plates.

From the above,

F = (mgd + MgD + mad + MAD)/b = (md(g+a) + MD(g+A))/b

Substituting actual numbers

F = (2.2×0.13×(9.81+0.9) + 1.1×0.37×(9.81+2.5))/0.035 = 230.7 N