APparent Weight (1 Viewer)

[GaganDeep]

I don't understand the apparent weight thing
Sum of F witha arrow to the right = ma (the "a" has a arrow at the top pointing to the right)
then it says "so that T-mg=ma"
T= mg+ma
The T's and g's and a's have the arrow.

Could some one please explain this.
I have no clue what it is.

Also this question
A model rocket has a pre launch mass or 85 g, of which 25g is solid propellant. It is able to deliver a trhust of 3.8N for a period of 2.8s. Assuming that hte rocket is fired directly up, determine
1) initial rate of accleration and g force
b) the final rate of acceleration and g force just prior to exhaustion of the fuel
In the solution for B) the "g" force
9.8 + 54
---------
9.8

the 54 being the final acceleration

My question is why are you adding 9.8 isn't it the gravity value at the specific altitude?, I know it's not significant but still.
Also the question states it's fired directly up, if it wasn't what would be different?
Thanks

 

[Riviet]

The arrow above all those units just means that the unit is a vector, which is a quantity that has both magnitude and direction. Some textbooks also use a squiggly line on top, which means the same thing.

 

[GaganDeep]

Thanks, could you please also explain the app weight thing. Thanks

Also this question
A model rocket has a pre launch mass or 85 g, of which 25g is solid propellant. It is able to deliver a trhust of 3.8N for a period of 2.8s. Assuming that hte rocket is fired directly up, determine
1) initial rate of accleration and g force
b) the final rate of acceleration and g force just prior to exhaustion of the fuel
In the solution for B) the "g" force
9.8 + 54
---------
9.8

the 54 being the final acceleration

My question is why are you adding 9.8 isn't it the gravity value at the specific altitude?, I know it's not significant but still.
Also the question states it's fired directly up, if it wasn't what would be different?
Thanks

 

[Riviet]

An object's weight, henceforth called "actual weight", is the downward force exerted upon it by the earth's gravity. By contrast, an object's apparent weight is the upward force (the normal force, or reaction force), typically transmitted through the ground, that opposes gravity and prevents a supported object from falling.

It is apparent weight, rather than actual weight, that a spring weighing scale measures. Apparent weight is also responsible for our sensation of the weight of our own bodies.

An object's apparent weight is equal to its actual weight, except if:

* The object has an acceleration (relative to the earth) with a vertical component, as in a lift, a rocket or a rollercoaster.
* Some force other than the earth's gravity and the normal force is acting on the object. This may be buoyancy, magnetic force, centrifugal force, or the gravitational force of another body.

For every action, there is an equal and opposite reaction force. The apparent weight is the reaction force to the actual weight.

 

[simbim21]

hey, i dunno if this is right (check your answers) but here's how i'd do it:

m= 85g (25g fuel)
T= 3.8N , time= 2.8s

a) F=ma
3.8= 0.085xa
:. a= 44.7m/s/s
:. g force= 44.7/9.8
= 4.6g

b) after 2.8 sec, m= 85-25= 60g
T= 3.8N

F=ma
3.8= .06xa
:.a= 63.3 m/s/s up

however, the rocket is going upwards while gravity is acting downwards. so the net acceleration = 63.3-9.8
= 53.5 m/s/s (~54)

however the g-forces are acting on the astronauts inside the rocket, who are experiencing an acceleration of 63.3m/s/s inside the rocket. hence, the gforces = 63.3/9.8
= 6.5g

that's why the answers are 54 + 9.8/9.8. I THINK...im really not so sure

 

[GaganDeep]

h hence, the gforces = 63.3/9.8
= 6.5g

that's why the answers are 54 + 9.8/9.8. I THINK...im really not so sure

But it won't be 9.8 at every altitude, that's what i am wondering. As the rocket goes the the gravity effect decreases, right? Wouldn't you calculate that and add that value to 54?

 

[Riviet]

But it won't be 9.8 at every altitude, that's what i am wondering. As the rocket goes the the gravity effect decreases, right? Wouldn't you calculate that and add that value to 54?

Yes, you're right that as you get further away from earth, gravity gradually decreases, but we still aren't far enough away from the earth for the distance to reduce the gravity by that much at this point of the flight.

 

[simbim21]

exactly. remember it is only a model rocket. so its not going to get very high, so change in gravity is negligible.
though if it were a question referring to a real rocket, then you would calculate gravity at that time (eg. 2.8s for this ques).

 

[zeropoint]

For every action, there is an equal and opposite reaction force. The apparent weight is the reaction force to the actual weight.

No, no, no! The apparent weight is not the reaction to the weight force! It is the normal force exerted by the chair to keep you accelerating, or otherwise stationary on the surface of the earth.

 

[Riviet]

Sorry, my bad, I was thinking of what you said as I referred to the apparent weight.