Finding mass of object based on torque

[Hans Herland]

Homework Statement

As the title states, the goal of this problem is to determine the weight of an object based on the torque caused by a force applied to it. The object in question is a wheelbarrow, and the force applied is a person lifting it by the handle using 100N with an angle of -30° from the x-axis. Hopefully the image I'm providing gives a sense of the problem.

The resultant torque caused by the force in point B combined with the torque caused by the weight of point G to point A is 0.

Sorry for completely ruining the scaling of distances, but the numbers are correct. =)

Homework Equations

For the torque equations so far we've been using: ΣM = F_xd_y - F_yd_x

For G, the equation is: G = mg. That is, at least the one I think would be relevant here.

The Attempt at a Solution

What I've attempted so far is to first calculate the torque caused by the applied force in point B. I am a little uncertain as to how the length of the arm works in that equation. Is, for instance, the length of arm B_x -1.2, or do you just use 1.2 without taking into consideration what direction the arm is?

(100 cos 30)(-0.65) - (100 sin 30)(-1.2) = 3.7Nm

So far, I feel like I'm on the correct path. With this answer, I start looking at the torque in point A, and that's where I start feeling unsure.

(F_Ax)(0.5) - (F_Ay)(0.3) = -3.7Nm
Here I have two unknowns, since I don't know the actual force on point A. What I've tried to do here is to replace those unknowns with the mass of point G, mg, but do I need to apply that with trigonometry?

The angle from point A to G is 59° when I use tan^-1 (0.5/0.3).

(mg cos 59)(0.5) - (mg sin 59)(0.3) = -3.7
m(9.81 cos 59(0.5)) - m(9.81 sin 59(0.3)) = -3.7
m(2.526 - 2.522) = -3.7
m = -3.7 / (2.526 - 2.522) = 925kg

I've done some other attempts, but I don't get as far as I feel I do here. 925kg doesn't sound like a reasonable answer, so there's something I'm not getting here.

I'd be grateful for any suggestions. =)

 

[haruspex]

(100 cos 30)(-0.65) - (100 sin 30)(-1.2) = 3.7Nm

Not sure how you are determining the signs there. Seems to me that both the horizontal and vertical components of the applied force act anticlockwise around G, so they should reinforce, not cancel. (It is so easy to get signs wrong that it is essential to do that sort of reality check.)

Here I have two unknowns

Since you appear to be taking moments about G, I assume these two forces are the forces the wheel exerts on the axle at A. That is a bit awkward because the wheel may have some mass. Simpler will be to take the wheel as part of the wheelbarrow, so consider forces at point of contact with ground. If we ignore any friction at the axle, and any moment of inertia of the wheel, we can write down straight away what the horizontal force must be there.

 

[JMatt7]

(I'm going to assume that the wheelbarrow is standing on the ground, and therefore the force acting in A is the reaction force from the ground)

First of all, if you want to apply the torque equation you have to choose a pivot point and then calculate the torques relative to that point.
The choice is completely arbitrary, but choosing the right point may help in solving the problem faster since it may be chosen to eliminate some unknown forces.
In your case, there are three possible points one could choose: A, B or G.

From what I understand, it seems you chose to calculate all torques relative to point G.
That's okay, but it's not the best choice since as you noticed you're left with an unknown force acting in point A.

However, this is not a big problem. As you should know there are two equations for static equilibrium: the toque equation $$∑ \vec M_i = 0$$ and the net force equation $$∑ \vec F_i = 0$$ so to calculate the components of the force in A, you have to use that second equation.

 

[haruspex]

you have to choose a pivot point and then calculate the torques relative to that point.
The choice is completely arbitrary,

That is true in statics, but clearly there is acceleration in the present problem.

 

[Hans Herland]

Not sure how you are determining the signs there. Seems to me that both the horizontal and vertical components of the applied force act anticlockwise around G, so they should reinforce, not cancel. (It is so easy to get signs wrong that it is essential to do that sort of reality check.)

I'll be honest, I'm not even sure myself how I determine the signs in some cases. I find it really confusing.

The problem I encounter is that when I look at the formula, F_xd_y - F_yd_x, and write it as:
(100 cos 30)(0.65) - (100 sin 30)(1.2) = -3.7
So I get a negative number when calculating the torque from A around point G, indicating that the force is directed clockwise since that formula is, from my understanding, supposed to set anticlockwise as the positive number. However, that feels wrong since lifting the wheelbarrow should give it a anticlockwise rotation from the torque generated at point A.

This is why it confuses me when it comes to selecting the signs in that formula.

Since you appear to be taking moments about G, I assume these two forces are the forces the wheel exerts on the axle at A. That is a bit awkward because the wheel may have some mass. Simpler will be to take the wheel as part of the wheelbarrow, so consider forces at point of contact with ground. If we ignore any friction at the axle, and any moment of inertia of the wheel, we can write down straight away what the horizontal force must be there.

I do notice that I did fail on one part of the drawing. There is a block in front of the wheel that does seem to signalize that the wheelbarrow isn't going to move anywhere. The problem is finding out the mass of the wheelbarrow, a mass which is centered at G.

Maybe it's better if I write the actual problem text, instead of trying to explain it myself?

The wheelbarrow and its contents have a center of mass at G. If F = 100 N and the resultant moment produced by force F and the weight about the axle at A is zero, determine the mass of the wheelbarrow and its contents.

I see that I should perhaps look at the force from point A as being directed straight up, as it's the reaction force from the ground that counters the torque from the force at point B. I'm guessing the block in front of the wheel that I failed to draw also counters the B_x force.

I'm just a little unsure as to exactly how I proceed here, seeing that I also have a hard time finding the proper torque generated at point B. Does -3.7Nm sound like a reasonable answer here? In my head it sounds to be going the opposite direction of what should be expected.

 

[haruspex]

Fxdy - Fydx

That assumes a consistent convention of positive directions. If we take up and right as positive here Fx will beb negative, but Fy, dx and dy will all be positive. So, as I indicated, the two terms will reinforce, not cancel.

The information about the block in front of the wheel certainly makes it conceptually simpler.
In view if the given information regarding net torque about A, it seems more natural to calculate all torques about A, not G. Did you try that?

 

[Hans Herland]

That assumes a consistent convention of positive directions. If we take up and right as positive here Fx will beb negative, but Fy, dx and dy will all be positive. So, as I indicated, the two terms will reinforce, not cancel.

I'm having a bit problems understanding exactly what this means, which is probably caused by me not being a native english speaker. Could you explain what "consistent convention of positive directions" means?

From what I understand from my textbook, we look at torque with the right hand rule, meaning we look at spins going anticlockwise as positive. So since the book shows the formula of torque calculation being F_xd_y - F_yd_x, then I assumed that getting a negative number in the equation indicated that the torque was going clockwise.

If we take up and right as positive here Fx will beb negative, but Fy, dx and dy will all be positive.

That makes sense, and I see that I haven't looked at it like that. If I write the equation setting F_x as negative, then I get the torque to be -174.6Nm. But can this number still indicate the torque being anticlockwise?

In view if the given information regarding net torque about A, it seems more natural to calculate all torques about A, not G. Did you try that?

That sounds way more obvious than looking at all torques about G like I've been doing. =)

I'll give this another go tomorrow, it's 2am here for me. Thank you so far for your help.

 

[haruspex]

we look at torque with the right hand rule, meaning we look at spins going anticlockwise as positive

Ok, but that must go together with other conventions. Presuming up and right as positive, the formula you quote, Fxdy - Fydx, only works if displacement is measured from force to axis, not the other way around. Using that, Fy is positive but Fx, dx and dy are negative. That gives you a positive torque.
I always take displacement as being from axis to force. Are you sure the formula is not Fydx-Fxdy?

 

[Hans Herland]

I've marked this as solved. I had a talk with my lecturer for help on how to solve it. It was much easier than I was making it, as always. =)

ΣMa = mg(0.3) - (100 cos30)(1.15) - (100 sin 30)(1.5) = 0
$$m = \frac {(100 cos30)(1.15) + (100 sin30)(1.5)} {(9.81)(0.3)} = 59.3kg$$

Thanks to everyone that helped out. =)

 

[haruspex]

I've marked this as solved. I had a talk with my lecturer for help on how to solve it. It was much easier than I was making it, as always. =)

ΣMa = mg(0.3) - (100 cos30)(1.15) - (100 sin 30)(1.5) = 0
$$m = \frac {(100 cos30)(1.15) + (100 sin30)(1.5)} {(9.81)(0.3)} = 59.3kg$$

Thanks to everyone that helped out. =)

Sure, but in terms of your original question, that is taking the torque to be F_yd_x-F_xd_y, not as in the formula you quoted.

 

[Hans Herland]

Sure, but in terms of your original question, that is taking the torque to be F_yd_x-F_xd_y, not as in the formula you quoted.

Yeah, I noticed I was misunderstanding what the book was saying. My lecturer showed me how it worked, and had a laugh when she saw the attempts I had done where I filled pages with calculations. =)