# Find Forces on Hand and Arm Holding a Ball

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1. Nov 1, 2014

### Gwozdzilla

1. The problem statement, all variables and given/known data
When an object is static (not moving), forces and moments must balance to zero. A moment here is defined as the force applied to an object multiplied by the distance from the point of rotation (torque). A simple way of thinking about moment balances is to consider a seesaw with a large child on one side and a small child on the other. The large child must sit loser to the fulcrum to balance the small child.

Using force and moment, balances, find the magnitude FA, FB, and Fc in the following diagram (attached below). What is actually applying the force FB?

2. Relevant equations

ΣFy = 0
ΣFx = 0
Σ T = 0
F = ma
T = Fd

1lb = 4.45N
1in = 2.54cm

3. The attempt at a solution

Well, first I converted all of my units from pounds to newtons and inches to meters. I assigned the variables wB for the weight of the ball and wA for the weight of the arm.

wB = 10lbs = 44.5N
wA = 3.5lbs = 15.58N

ΣFy = 0
FBy - FA = -wB - wA
FBy = FBsin(75)
FBsin(75) - FA = -wB - wA

ΣFy = 0
FC = -FBx

Torque from arm weight = wAdA = (15.58N)(.183m)
Torque from ball = wBdB = (44.5N)(.356m)
Visually, these two torques can't possibly sum to zero, so I must be missing a torque...

I know that the next step is to solve a system of equations, but I'm pretty sure these aren't the correct equations because I don't know how to draw a free body diagram for this complicated of a situation with all of the forces acting at different distances. I guess my questions are what forces am I missing? and when I draw the free body diagram, do I draw it as if all of the forces are acting at the same point?

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Last edited: Nov 1, 2014
2. Nov 2, 2014

### haruspex

Check the signs.
There are two other forces. What torques do they have about the pivot point?

3. Nov 2, 2014

### Gwozdzilla

To try to correct the signs:

FBy is actually in the opposite direction of all the other forces, so I guess the formula is FBy = FA + wB + wA ?

TFB = FByd = FB(sin(75))(.0508m)

TFA = FAd = FA 0 because the force is acting at a distance 0 from the rotation?

If these are the two additional torques, then using the same logic I used to fix the signs in the ∑Fy equation, the torques would sum like this:

TFB = TwB + TwA

Are these adjustments correct? If I take my three formulas ΣFy, ΣFx, and ΣT, can I solve the problem as a system of equations and get the correct results?

I don't think it matters much to the solution of this problem (it might have something to do with figuring out what's applying the force FB), but where are FA and FC coming from? Are they normal forces from something? Had I not been given the diagram, I wouldn't have thought that they were there at all.[/SUB]

4. Nov 2, 2014

### BvU

FBy is actually in the opposite direction of all the other forces, so I guess the formula is FBy = FA + wB + wA ?

It is better to stick to $\displaystyle \Sigma \vec F = 0$ and assign the proper sign to each of the forces.
Since wB and wA are pointing in the negative y-direction, you get something like
FBy + FA + wB + wA = 0 $\Leftrightarrow$
FBy = $\bf -$ ( FA + wB + wA ) with all three forces on the righthand side < 0 .
TFB = FByd = FB(sin(75))(.0508m) Yes

Actually, there's a bit more to this: torque is a pseudo vector and it's defined as a cross vector product $\vec \tau \equiv \vec r \times \vec F$. Since, in a right-handed coordinate system $\hat z = \hat x \times \hat y$, you indeed get a torque in the positive z-direction (r is pointing in the postive x, FB in the positive y direction). As with the forces: this means the torques from wA and wB are < 0 in ΣT = 0

If you're not into cross-products yet: distinguish torque wanting a clockwise rotation from torque wanting an anti-clockwise rotation by giving the former a minus sign. That way you can still require ΣT = 0 for equilibrium (no angular acceleration)

TFA = FAd = FA 0 because the force is acting at a distance 0 from the rotation? Yes

It doesn't matter to the solution, I agree. But it greatly helps towards understanding if you can figure out where, in particular, FB comes from. What do you do when you lift a weight that way ?

FA shows up from the force balance: if it weren't there, your arm would be pulled up. It isn't so there must be something holding it back at the elbow...

5. Nov 2, 2014

### Gwozdzilla

Maybe I'm confused about whether the forces are assumed to be negative or positive when they're listed as variables. Are wA, wB, and FA assumed to have negative values because they're in the -y direction?

Doing what I can to avoid the variables, is this correct?
If...
|wB| = 44.5N
|wA| = 15.58N
and FA is assumed to have a negative value because of its direction, then...

FBsin(75) + FA - 15.58N - 44.5N = 0

The same could follow with the torques if the values of the variables for the torques going in the clockwise direction are assumed to be negative, then..

ΣT = TwA + TwB + TFB = 0

How does the arm get pulled upward without FA? Is FA supposed to be due to the weight of the upper arm? And it might be a really poor guess, especially with how the diagram is labeled, but is FB caused by the muscles in the forearm? FB is the only force preventing the whole system from falling down due to gravity. I'm not really sure how to figure this part out.

6. Nov 2, 2014

### BvU

Maybe I'm confused about whether the forces are assumed to be negative or positive when they're listed as variables. Are wA, wB, and FA assumed to have negative values because they're in the -y direction? Yes. Magnitude is an absolute value, hence always positive, value is negative.

Doing what I can to avoid the variables, is this correct?
If...
|wB| = 44.5N Yes
|wA| = 15.58N Yes
and FA is assumed to have a negative value because of its direction, then...

FBsin(75) + FA - 15.58N - 44.5N = 0 Yes

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Found out what causes FB already? If not, feel between shoulder and elbow with one hand while holding up a suitable weight in the other..... If still no clue, compare with no weight situation (wA = 0), if so desired laying hand on table ( (wB = 0, too ). -- Funny: Too) becomes a smiley
If all that doesn't work: try to lift a table or something still heavier. ;)

FB from the torques should give a considerable magnitude, such that FBsin(75) - 15.58N - 44.5N >> 0 : if there were no FA, the arm would move up. Something solid is in the way. See your picture (top one)...

7. Nov 2, 2014

### haruspex

It's up to you. You can pick a direction (e.g. up) as positive and accept that some will take negative values, or you can take each vector to be positive in the direction of the arrow in your diagram (yet some may still turn out negative if you guessed wrongly). The trouble is, you need to state what you are doing if others are to understand your equations.