### Video Transcript

The forearm shown below is
positioned at an angle π with respect to the upper arm. And a 5.0-kilogram mass is held in
the hand. The total mass of the forearm and
hand is 3.0 kilograms. And their center of mass is 15.0
centimeters from the elbow. What is the magnitude of the force
that the bicep muscle exerts on the forearm for π equals 60 degrees? What is the magnitude of the force
on the elbow joint for a π equals 60 degrees?

We can label the force magnitude of
the bicep πΉ sub π΅ and the force magnitude on the elbow πΉ sub π. When we consider this scenario, we
have a ball being held motionless. Thereβs no rotation and thereβs no
translation going on. If we consider the forearm of this
arm as the length along which torques and forces are exerted, we can say that the
sum of all the torques exerted on the forearm are zero. We will use this condition along
with the other information weβve been given about the scenario to solve for the
force exerted by the bicep, πΉ sub π΅. We can simplify this scenario by
modeling the forearm simply as a rigid extended line segment.

When it comes to the forces that
are acting on this forearm, we know thereβs the weight force of the five-kilogram
ball acting down. Weβll label that force π sub π,
the mass of the ball of 5.0 kilograms, times π, the acceleration due to gravity,
which is 9.8 meters per second squared. Weβre also told about the mass of
the arm, that is, the forearm and hand combined. That mass also has a downward force
that is exerted on this line. Weβll call that force π sub π for
the mass of the arm, given as 3.0 kilograms times π.

We also know that the bicep muscle
exerts an upward force on the forearm in order to help support the weight of the
ball. This is the force we want to solve
for. And weβre presuming that these
three forces are acting around a rotation axis at the elbow joint. Our next step is to calculate the
torque created about this rotation point at the elbow joint by each of these three
forces. Recalling that torque magnitude is
equal to force times distance from the axis of rotation multiplied by the sine of
the angle between these two vectors, and since weβre told that, in our case, the
angle π, which is also the same as the π shown in this general equation, is equal
to 60 degrees. Weβre almost ready to write out the
torques created by these three forces acting on the forearm.

But before we do, letβs choose
which direction of rotation about our rotation point we want to consider
positive. Weβll choose clockwise rotation as
positive rotation. This choice means that π sub π
times π and π sub π times π will create positive torques and πΉ sub π΅, our
bicep force, will create a negative torque. When we enter the expression πΉ
times π times the sin of π for each one of the three forces acting on our forearm
and set that sum equal to zero, we use the fact that the mass of the arm is said to
have a center of gravity 15.0 centimeters from the elbow. And the ball is 35.0 centimeters
from the elbow joint, while the bicep is only 4.0 centimeters away.

Since the sin of 60 degrees is
common to all the terms on the left-hand side of our equation, we can divide both
sides of the equation by that term and cancel it out. Now since we want to solve for πΉ
sub π΅, we rearrange to isolate that on one side of the equation. When we do and also factor out the
acceleration due to gravity π, we see that our expression for πΉ sub π΅ is in terms
of known values π sub π, π sub π, and our distances, as well as π. Weβll now plug in for those masses
as well as convert our distances from units of centimeters to meters. When we enter this expression on
our calculator, we find that, to two significant figures, πΉ sub π΅ is 540
newtons. Thatβs the force exerted by the
bicep on the forearm in order to ensure rotational equilibrium.

The next thing we wanna do is solve
for the force thatβs exerted on the elbow joint. Now you might say, βWhat force? I thought we were in
equilibrium.β Itβs true that weβre in rotational
equilibrium with the three forces weβve named about the axis weβve chosen at the
elbow joint. But the equilibrium weβll use to
solve for πΉ sub π is a translational equilibrium. That is, the whole arm isnβt moving
up, down, left, or right at all. The implication is that not only is
the sum of torques on this forearm zero, but the sum of the forces on the forearm is
zero as well. If we look back at our original
diagram and specifically at the elbow joint, if we had to guess a direction for the
force exerted on that joint because of the rest of this scenario. We could say that because bones
tend to exert force while under compression that that elbow force would point in the
same direction as the weight force on the ball and the arm.

If we decide to set up the
convention that downward direction is positive force, then because all the forces on
the forearm add up to zero, we can write that πΉ sub π plus π times the quantity
π sub π plus π sub π minus πΉ sub π΅ is equal to zero. Or πΉ sub π is equal to πΉ sub π΅
minus π times the quantity π π plus π π. Weβve been given the values for π
sub π and π sub π. We know π since itβs a
constant. And we solved for πΉ sub π΅ in a
previous part. So weβre ready to plug in and solve
for πΉ sub π. When we enter this expression on
our calculator, again to two significant figures, we find that πΉ sub π is 460
newtons. Thatβs the force magnitude exerted
on the elbow joint.