Mass on an incline OT

Discussion in 'NZ Computing' started by Acetyldehyde, Feb 26, 2004.

1. AcetyldehydeGuest

When I push 100 Kg straight up it weighs 100kg.

When I push 100 Kg up a 45 deg. slope (assume 0.0 friction) how much does it
weigh?

Acetyldehyde, Feb 26, 2004

2. PeterGuest

Acetyldehyde wrote:

> When I push 100 Kg straight up it weighs 100kg.
>
> When I push 100 Kg up a 45 deg. slope (assume 0.0 friction) how much does
> it weigh?

the weight component down the slope is
100 * SIN(45)
= 71 kg

When you're calculating this in a spreadsheet or calculator, check whether
the SIN function works in degrees or radians.

HTH

Peter

Peter, Feb 26, 2004

3. Robert CoozeGuest

Acetyldehyde wrote:
> When I push 100 Kg straight up it weighs 100kg.
>
> When I push 100 Kg up a 45 deg. slope (assume 0.0 friction) how much does it
> weigh?
>
>

LOL Go to school and do 6th form physics

answer it will still weigh the same

I think the question should be along the lines of

How much force
How much power
How much work

In answer if you use the same amount of effort and 1/2 the time you will
have used about 1/2 the power
Or use 1/2 the force and the same time you will have used 1/2 the power

also weight can be measured in mass or force mass changes in gravity
force does not ( I may have totally got this wrong ) its been 26 years
since I have had anything to do with 6th form physics

Let the flame War start

......................"""""",,,,,''''''??? (please insert the punctuation
as you see fit)

--
http://cooze.co.nz home of the RecyclerMan aka Robert Cooze

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Robert Cooze, Feb 26, 2004

On Fri, 27 Feb 2004 07:31:03 +1300, Robert Cooze wrote:

> also weight can be measured in mass or force mass changes in gravity force
> does not ( I may have totally got this wrong )

Well you got the last bit right

Weight is a force due to gravity measured in newtons, mass is a constant
property of the object measured in kg. Roughly speaking, mass is related
to inertia ie how much effort is required to accelerate something - I
forget the exact distinction between mass and inertia

And as for what the original poster was asking about how does weight
change when pushing something up a slope, well depending on what they
really were asking, the weight doesn't change (ignoring any significant
changes in gravitational field by going really really high), but the force
required to hold it on the slope is the component of weight down the slope
(as Peter mentioned).

Only 16 years since 6 form physics

Cheers
Anton

AD., Feb 26, 2004
5. T-BoyGuest

In article <403e4f38\$>, says...
> Acetyldehyde wrote:
> > When I push 100 Kg straight up it weighs 100kg.
> >
> > When I push 100 Kg up a 45 deg. slope (assume 0.0 friction) how much does it
> > weigh?
> >
> >

> LOL Go to school and do 6th form physics
>
> answer it will still weigh the same

No it won't. It's mass will be the same. It's weight (a measure of
*force* - will not be the same). (As you managed to correctly conclude
later on)

--
Duncan

T-Boy, Feb 26, 2004
6. KeithGuest

T-Boy <> wrote:
>In article <403e4f38\$>, says...
>> Acetyldehyde wrote:
>> > When I push 100 Kg straight up it weighs 100kg.
>> >
>> > When I push 100 Kg up a 45 deg. slope (assume 0.0 friction) how much does it
>> > weigh?
>> >
>> >

>> LOL Go to school and do 6th form physics
>>
>> answer it will still weigh the same

>
>No it won't. It's mass will be the same. It's weight (a measure of
>*force* - will not be the same). (As you managed to correctly conclude
>later on)

For a graphical representation (and lots of other physics stuff) see:
http://www.walter-fendt.de/ph11e/inclplane.htm

Keith, Feb 26, 2004
7. Steve RobertsonGuest

>No it won't. It's mass will be the same. It's weight (a measure of
>*force* - will not be the same). (As you managed to correctly conclude
>later on)
>

If you put a scale under on the incline surely it would weigh the same??
but would require less force (but the same toatal energy?) to push it.

Isnt this like the 'theory' that if you weigh yourself with scales on a shag
pile capert it will give a lower reading.

Steve Robertson, Feb 27, 2004