Video: Analyzing Forces and Moments in a System in Equilibrium

Video Transcript

The forearm shown below is positioned at an angle 𝜃 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 𝜃 equals 60 degrees? What is the magnitude of the force on the elbow joint for a 𝜃 equals 60 degrees?

We can label the force magnitude of the bicep 𝐹 sub 𝐵 and the force magnitude on the elbow 𝐹 sub 𝑒. When we consider this scenario, we have a ball being held motionless. There’s no rotation and there’s 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 we’ve been given about the scenario to solve for the force exerted by the bicep, 𝐹 sub 𝐵. 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 there’s the weight force of the five-kilogram ball acting down. We’ll label that force 𝑚 sub 𝑏, the mass of the ball of 5.0 kilograms, times 𝑔, the acceleration due to gravity, which is 9.8 meters per second squared. We’re 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. We’ll call that force 𝑚 sub 𝑎 for the mass of the arm, given as 3.0 kilograms times 𝑔.

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 we’re 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 we’re told that, in our case, the angle 𝜃, which is also the same as the 𝜃 shown in this general equation, is equal to 60 degrees. We’re almost ready to write out the torques created by these three forces acting on the forearm.

But before we do, let’s choose which direction of rotation about our rotation point we want to consider positive. We’ll choose clockwise rotation as positive rotation. This choice means that 𝑚 sub 𝑎 times 𝑔 and 𝑚 sub 𝑏 times 𝑔 will create positive torques and 𝐹 sub 𝐵, our bicep force, will create a negative torque. When we enter the expression 𝐹 times 𝑟 times the sin of 𝜃 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 𝐹 sub 𝐵, we rearrange to isolate that on one side of the equation. When we do and also factor out the acceleration due to gravity 𝑔, we see that our expression for 𝐹 sub 𝐵 is in terms of known values 𝑚 sub 𝑎, 𝑚 sub 𝑏, and our distances, as well as 𝑔. We’ll 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, 𝐹 sub 𝐵 is 540 newtons. That’s 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 that’s exerted on the elbow joint. Now you might say, “What force? I thought we were in equilibrium.” It’s true that we’re in rotational equilibrium with the three forces we’ve named about the axis we’ve chosen at the elbow joint. But the equilibrium we’ll use to solve for 𝐹 sub 𝑒 is a translational equilibrium. That is, the whole arm isn’t 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 𝐹 sub 𝑒 plus 𝑔 times the quantity 𝑚 sub 𝑎 plus 𝑚 sub 𝑏 minus 𝐹 sub 𝐵 is equal to zero. Or 𝐹 sub 𝑒 is equal to 𝐹 sub 𝐵 minus 𝑔 times the quantity 𝑚 𝑎 plus 𝑚 𝑏. We’ve been given the values for 𝑚 sub 𝑎 and 𝑚 sub 𝑏. We know 𝑔 since it’s a constant. And we solved for 𝐹 sub 𝐵 in a previous part. So we’re ready to plug in and solve for 𝐹 sub 𝑒. When we enter this expression on our calculator, again to two significant figures, we find that 𝐹 sub 𝑒 is 460 newtons. That’s the force magnitude exerted on the elbow joint.