physics problem

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sv3

Full Member
10+ Year Member
This question was posted in the MCAT section but i know the problem solving fiends reside here.......I solved this using intuition, but can't solve it using math that makes sense to me. I can play with the numbers till I get the answer but not in a way that makes sense so hoping someone could shed light on it for me......

"If a 50 kg woman on a scale is in an elevator initially going up at 20 m/s then 2 seconds later is going up at 10 m/s what is her apparent weight?"

thanks

Plasma1113

New Member
10+ Year Member
This question was posted in the MCAT section but i know the problem solving fiends reside here.......I solved this using intuition, but can't solve it using math that makes sense to me. I can play with the numbers till I get the answer but not in a way that makes sense so hoping someone could shed light on it for me......

"If a 50 kg woman on a scale is in an elevator initially going up at 20 m/s then 2 seconds later is going up at 10 m/s what is her apparent weight?"

thanks

50kg*g=500N=Force of gravity on object
a=(10m/s-20m/s)/2s=-5m/s^2 The object is slowing down.

50kg*a=-250N=Force of acceleration on object.

The force of gravity on the object and the force of acceleration on the object are acting in opposite directions; this is a key aspect of this problem.

Add because the direction of the force is already taken into account with the signs (+/-).
(If direction wasn't taken into account you would subtract in this case.)

500N+(-250N)=Net Force=250N

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Fort

Full Member
10+ Year Member
We can use Newton's Second Law to solve this. When she slows down she should feel lighter (from your own experience this is true as you are nearing your desired floor), so she should weigh less than 500N. We can draw a free body diagram of the elevator and notice there are two forces acting on her: Normal and gravity.

So from Newton's Second Law we have

F(net) = ma

F(normal)-F(gravity) = ma (I've defined up as positive)

We want to solve for the normal force, but first we need to calculate here acceleration during this time frame.

a = v_2-v_1/t

a = (10m/s-20m/s)/2s

a = -5m/s^2

So we see she is decelerating as she nears her stop!

Returning to our force equation we have

F(normal) = ma + F(gravity)

F(normal) = m(g+a)

F(normal) = (50kg)*(10m/s^2-5m/s^2)

F(normal) = 50kg(5m/s^2)

F(normal) = 250 N

Hope that helps.

sv3

Full Member
10+ Year Member
We can use Newton's Second Law to solve this. When she slows down she should feel lighter (from your own experience this is true as you are nearing your desired floor), so she should weigh less than 500N. We can draw a free body diagram of the elevator and notice there are two forces acting on her: Normal and gravity.

So from Newton's Second Law we have

F(net) = ma

F(normal)-F(gravity) = ma (I've defined up as positive)

We want to solve for the normal force, but first we need to calculate here acceleration during this time frame.

a = v_2-v_1/t

a = (10m/s-20m/s)/2s

a = -5m/s^2

So we see she is decelerating as she nears her stop!

Returning to our force equation we have

F(normal) = ma + F(gravity)

F(normal) = m(g+a)

F(normal) = (50kg)*(10m/s^2-5m/s^2)

F(normal) = 50kg(5m/s^2)

F(normal) = 250 N

Hope that helps.

Thanks alot for the explanation. For some reason I am still having issues wrapping my head around the fact that the Force of gravity and the Force of acceleration are opposite to each other. In your response you said up was positive, but you are using a positve value for g. I thought this would be negative? Of course you get the wrong answer this way and i avoided this using the real life example you pointed out. So I'm still trying to solve this with math.......sorry for the trouble as i know this is an easy one but still managing to irk me. I know the way you did was spot on b/c it also asnwers the reverse question if she was accelerating.

cheers

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sv3

Full Member
10+ Year Member
50kg*g=500N=Force of gravity on object
a=(10m/s-20m/s)/2s=-5m/s^2 The object is slowing down.

50kg*a=-250N=Force of acceleration on object.

The force of gravity on the object and the force of acceleration on the object are acting in opposite directions; this is a key aspect of this problem.

Add because the direction of the force is already taken into account with the signs (+/-).
(If direction wasn't taken into account you would subtract in this case.)

500N+(-250N)=Net Force=250N

And that's precisely what is bugging me. If she is deccelerating, isn't that pointing in the same direction as gravity? (down) Now I know intuitively this makes no sense but I can't logically figure it out.

Fort

Full Member
10+ Year Member
Thanks alot for the explanation. For some reason I am still having issues wrapping my head around the fact that the Force of gravity and the Force of acceleration are opposite to each other. In your response you said up was positive, but you are using a positve value for g. I thought this would be negative? Of course you get the wrong answer this way and i avoided this using the real life example you pointed out. So I'm still trying to solve this with math.......sorry for the trouble as i know this is an easy one but still managing to irk me. I know the way you did was spot on b/c it also asnwers the reverse question if she was accelerating.

cheers

The force of gravity is pointed downwards meaning it's negative. This I have defined in my force equation. When dealing with gravity and forces, g should always be plugged in as a positive value, but the force that it provides may be negative depending on your sign conventions. Take a look at the equation below.

F(normal) - F(gravity) = ma (notice the minus sign because the forces are in opposing directions, ignore the acceleration term for now).

or more simply

F(normal) - mg = ma

See the weight vector is opposite the normal vector, so a positive value for g will suffice. The sign for the acceleration simply depends on if she's slowing down or not, which she is. Thus her acceleration vector is pointed in the same direction as mg and should be negative (which it is because we found it to be -5m/s^2, so all we do is plug this value in and the rest takes care of itself). When you're doing problems that involve vectors, i.e. force problems, make sure you define a coordinate system and everything will work itself out in the end. If I was told that she was decelerating in the elevator at say 7m/s^2, then that indicates the acceleration vector is pointing down, so I'd need to write my force equation like

F(normal) - F(gravity) = -ma (here there is a negative on the right side of the equation because I know that the acceleration vector is down and I simply plug in a value of 7m/s^2 for a),

It's confusing I know, but drawing a free body diagram helps a lot when identifying positive and negative forces on a body.

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sv3

Full Member
10+ Year Member
The force of gravity is pointed downwards meaning it's negative. This I have defined in my force equation. When dealing with gravity and forces, g should always be plugged in as a positive value, but the force that it provides may be negative depending on your sign conventions. Take a look at the equation below.

F(normal) - F(gravity) = ma (notice the minus sign because the forces are in opposing directions, ignore the acceleration term for now).

or more simply

F(normal) - mg = ma

See the weight vector is opposite the normal vector, so a positive value for g will suffice. The sign for the acceleration simply depends on if she's slowing down or not, which she is. Thus her acceleration vector is pointed in the same direction as mg and should be negative (which it is because we found it to be -5m/s^2, so all we do is plug this value in and the rest takes care of itself). When you're doing problems that involve vectors, i.e. force problems, make sure you define a coordinate system and everything will work itself out in the end. If I was told that she was decelerating in the elevator at say 7m/s^2, then that indicates the acceleration vector is pointing down, so I'd need to write my force equation like

F(normal) - F(gravity) = -ma (here there is a negative on the right side of the equation because I know that the acceleration vector is down and I simply plug in a value of 7m/s^2 for a),

It's confusing I know, but drawing a free body diagram helps a lot when identifying positive and negative forces on a body.

That bolded part above is what i was exactly having issues with. I kept on thinking, if she is deccelerating, isn't that Force in the same direction as the Force of gravity? I guess adding them is double counting. Frankly I'm still trying to sort this out now. But at least I can solve this now mathematically......that was a great explanation. Thanks alot for the time!

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Dr Gerrard

Full Member
10+ Year Member
I dont know, the way I thought of it was in terms of relative acceleration.

If acceleration due to gravity is 10 m/s^2 down, but the person is also accelerating at 5 m/s^2 in the same direction, the person will only feel a net of 5 m/s^2. Thus, W = ma = 250 N.

EDIT: To see it mathematically, remember that the way we measure weight is by measuring the normal force.

Fn + mg = ma

"a" and "g" are both in the same direction, so lets say they are both negative.

Fn - mg = -ma

Fn = mg - ma

Fn = 500 - 250 (since a is negative 5 m/s/s)

This makes sense, because if acceleration was 0, then normal force would equal mg, which is correct.

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IncuSpy

Full Member
10+ Year Member
15+ Year Member
I used to have alot of trouble visualizing apparent weight problems until I thought of a good schematic. You normally feel your weight (full gravity) when standing still, because F normal = F gravity. Weightlessness is when part of your F normal (or all of it) is countered by something else.

That is, standing F normal = F gravity. If apparent weight decreases (becomes less than F gravity) you experience weightlessness.

Think about taking an elevator that accelerates fast upwards. F up takes a portion out of F normal (force of elevator up takes up part of your F normal). So F up + F normal = F gravity.

F up (ma = 50 x (20- 10 / 2)) + F normal (apparent weight) = F grav (mg)

250 + F normal = 500

F normal = apparent weight = 250