Mechanics Question (Balancing)

[Graffiti]

Here's the questions:

1.) A uniform plank of mass 12 kg and length L is positioned horizontally, with its two ends supported by sensitive scales. An object of mass 3 kg is placed a distance L/3 from the left end of the plank. What weight does the right-hand scale read?

2.) In the preceding question, what weight does the left-hand scale read?

I didn't understand the books explanation for this. If anyone could explain how to approach this problem, it'd be a huge help! Thanks in advance.

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[stee1erfan]

Here Weight and Mass are two diff things. you will have to find Weight of the plank first. Assuming that its on earth you can just approximate F=ma where m=12kg and a = 10m/s^2 which is -120 Newtons. so without the object placed the WEIGHT will be divided in half between the two ends which is -60 Newtons each. long story short, the object is placed close to the left end of the plank so the right end's WEIGHT overall will be less than the left. So 2/3 of the 3kg weight is on the left so 2/3 of 30 Newtons is 20 Newtons. So 60 + 20 = 80 Newtons on the left and 60 + 10 = 70 Newtons on the Right side. IS THAT CORRECT?

 

[Graffiti]

Here Weight and Mass are two diff things. you will have to find Weight of the plank first. Assuming that its on earth you can just approximate F=ma where m=12kg and a = 10m/s^2 which is -120 Newtons. so without the object placed the WEIGHT will be divided in half between the two ends which is -60 Newtons each. long story short, the object is placed close to the left end of the plank so the right end's WEIGHT overall will be less than the left. So 2/3 of the 3kg weight is on the left so 2/3 of 30 Newtons is 20 Newtons. So 60 + 20 = 80 Newtons on the left and 60 + 10 = 70 Newtons on the Right side. IS THAT CORRECT?

Yes, those answers are correct but could you elaborate more with your explanation. I'm still having trouble understanding this.

 

[collegestud2013]

Here's the questions:

1.) A uniform plank of mass 12 kg and length L is positioned horizontally, with its two ends supported by sensitive scales. An object of mass 3 kg is placed a distance L/3 from the left end of the plank. What weight does the right-hand scale read?

2.) In the preceding question, what weight does the left-hand scale read?

I didn't understand the books explanation for this. If anyone could explain how to approach this problem, it'd be a huge help! Thanks in advance.

1) you know the objects are in static equilibrium so the sum of the torques = 0. The easiest way to do this would be to arbitrarily treat the left end of the plank as the pivot, and making sure to follow a consistent sign convention (I'll use CCW as + and CW as -). , Note that the torque due to the weight of the plank will act at in the middle and in the CW direction, that the torque do to the hanging block will also act as CW, and the force DUE TO the right scale will act CCW. So summing the torques gives (Wl and Wr are the readings of the left and right scale, respectively):

Wl(0) - 30(L/3) -120(L/2) + Wr(L) = 0
-10L - 60L + Wr(L) = 0
(-10-60 +Wr)L = 0
Wr = 70 N

2) Since the two scales have to exactly support the combined weights of both objects, and now that you know Wr is 70 N you can just get Wl by subtracting Wr from the total weight of the plank and block:

Wl = 120 + 30 - Wr = 150 - 70 = 80 N

 

[Graffiti]

1) you know the object is in static equilibrium so the sum of the torques = 0. The easiest way to do this would be to treat the left end as the pivot, and making sure to follow a consistent sign convention (I'll use CCW as + and CW as -). , Note that the torque due to the weight of the plank will act at in the middle and in the CW direction, that the torque do to the hanging block will also act as CW, and the force DUE TO the right scale will act CCW. So summing the torques gives (Wl and Wr are the readings of the left and right scale, respectively):

Wl(0) - 30(L/3) -120(L/2) + Wr(L) = 0
-10L - 60L + Wr(L) = 0
(-10-60 +Wr)L = 0
Wr = 70 N

2) Since the two scales have to exactly support the combined weights of both objects:

Wl = 120 + 30 - Wr = 150 - 70 = 80 N

Ah, I see what you did. But what I'm confused about is - why did you assume the weights of the left and right scale were both acting CCW.

 

[collegestud2013]

CCW and CW terminology can only work if there's some reference point, which I arbitrarily designated to be the left end of the plank since that's the easy way to do things based on whats given and what you need to find. So with the left end of the plank as the pivot, the force due to the left scale would point up but in this case it (again, based on my arbitrary designation) would neither be CCW nor CW since it's directly acting the pivot. And since the torques due to the weights of the block and plank are to the right of the pivot and point down, that would be make them CW (try visualizing or drawing this if you don't see it right away). And once you know that, then you know the torque due to the right scale has to point up and CCW in order to cancel out the weight torques.

 

[Graffiti]

CCW and CW terminology can only work if there's some reference point, which I arbitrarily designated to be the left end of the plank since that's the easy way to do things based on whats given and what you need to find. So with the left end of the plank as the pivot, the force due to the left scale would point up but in this case it (again, based on my arbitrary designation) would neither be CCW nor CW since it's directly acting the pivot. And since the torques due to the weights of the block and plank are to the right of the pivot and point down, that would be make them CW (try visualizing or drawing this if you don't see it right away). And once you know that, then you know the torque due to the right scale has to point up and CCW in order to cancel out the weight torques.

Ahhhhhhhh, I get it now! Thanks so much for your explanation. One question though, for "Center of Mass/Gravity" types of questions, do you normally balance out Torques to find the Balance Point? Does using torque (as opposed to the COM equation) normally work for these types of problems?

 

[collegestud2013]

no problem dude. I guess you could use torque concepts to solve certain center of mass questions but if it were on the MCAT I'd just go with what's faster and more familiar, which for me would just be the COM equation.