# Apparent Weight help!

1. Oct 19, 2005

### physixnot4me

Hi! I'm beginning a whole new unit in physics, that i'm not that sure of. Any suggestions would be appreciated as to how to tackle these following questions:

1) the pilot of an airplane executes a constant-speed loop the loop maneuver in a vertical circle. the speed of the airplane is 300mi/h, and the radius of the circle is 1200 ft.
(a) what is the pilot's apparent weight at the lowest point if his true weight is 160lb?
(b) what is his apparent weight at the highest point?
(note: his apparent weight is equal to the force exerted by the seat on his body)

2) a small container of water is placed on a carousel inside a microwave oven, at a radius of 12.0 cm from the center. The turntable rotates steadily, turning through one revolution in each 7.25s. What angle does the water surface make with the horizontal?

3) a person stands on a scale in the elevator. as the elevator starts, the scale has a constant reading of 591N. as the elevator later stops, the scale reading is 391N. assume the magnitude of the acceleration is the same starting and stopping, and determine:
(a) the weight of the person (b) the person's mass (c) the acceleration of the elevator

THANKS!

2. Oct 19, 2005

### Diane_

Generally speaking, when you're dealing with apparent weight, you're dealing with two things - gravity and some other force that either acts in concert with or in opposition to gravity. For instance, in your first problem, the actual weight of the pilot remains constant. When he's moving in the loop, though, the pilot's seat has to exert a centripetal force to keep him there. At the top of the loop, gravity is acting towards the center (which is pretty much the definition of "centripetal"), so the seat doesn't have to push as hard, if you will. At the bottom of the loop, gravity is acting as a centrifugal force (one of the few times you will actually see one of those), so the seat has to push not only to provide the entire centripetal force but also to counteract the pilot's weight. The force the seat provides, then, goes from a minimum at the top to a maximum at the bottom. Do you see how to proceed with that one?

The same sort of thing applies to the other two problems. Think about them and see if you can figure out exactly how.

Does that help?

3. Oct 20, 2005

### andrevdh

physixnot4me,
since the pilot is moving in a circe with a constant sped we know that he is accelerating towards the centre of the circle. The resultant force that he experiences must therefore always point towards the centre of the circle. This magnitude of this force can also be calculated from the given data. what you need to do is to draw a force diagram of the pilot and equate the resultant force to this centripetal force, since all the forces added up togehter must supply the centripetal force.

4. Oct 20, 2005

### andrevdh

The first attachment shows across section through a planet's interior. As one moves towards the center of the planet the pressure increases due to the compression of the matter by the weight of the outer material. In the second attachment a container of water is spun around it's center. Here we find that the pressure is a minimum at the centre and a maximum at the outer regions (how do we know this?). What causes this difference in pressure as we move to outer regions? How is this pressure difference generated?

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5. Oct 20, 2005

### physixnot4me

Just wondering how...

for these two questions:

2) a small container of water is placed on a carousel inside a microwave oven, at a radius of 12.0 cm from the center. The turntable rotates steadily, turning through one revolution in each 7.25s. What angle does the water surface make with the horizontal?

3) a person stands on a scale in the elevator. as the elevator starts, the scale has a constant reading of 591N. as the elevator later stops, the scale reading is 391N. assume the magnitude of the acceleration is the same starting and stopping, and determine:
(a) the weight of the person (b) the person's mass (c) the acceleration of the elevator

in #2, i'm stuck on how you would use the revolution/s within newtons's laws to find the angle at which the water surface would make with the horizontal.. what do they even mean by HORIZONTAL?

in #3, so the force drops from 591N to 391N (does this mean the elevator is going down... the apparent weight is becoming smaller?) and weight=mg? so would you solve that by subtracting the force when it stopped, from the starting force(total force) =mg, find Mass hat way?

6. Oct 20, 2005

### Staff: Mentor

Horizontal means what it usually means. If the carousel wasn't turning, the water level would be horizontal, just like usual. Hint: Find the acceleration, then find the direction of apparent gravity. (The surface of the water will be perpendicular to the apparent gravity.)
When the elevator starts, it accelerates upward; when it stops, it accelerates downward. Apply Newton's 2nd law to both cases.

7. Oct 21, 2005

### andrevdh

Doc Al,
why would the centrifugal acceleration be away from the centre, since centripetal acceleration is normally towards the centre?

8. Oct 21, 2005

### Staff: Mentor

The "smartass" answer is that the word centrifugal means "away from the center" and the word centripetal means "toward the center". (Sorry... couldn't resist. )

The real answer is that you can look at this problem from either of two reference frames:
(1) If you view things from the rotating (non-inertial) frame, then there appears a centrifugal force (acting away from the center). Thus the apparent weight will be the sum of $mg$ acting down plus $ma_c$ acting outward.

(2) You can also view things from the usual inertial frame. Apply Newton's 2nd law to an element of the water surface, realizing that in this frame the water undergoes a centripetal acceleration.​

9. Oct 21, 2005

### physixnot4me

just a question...

Sorry, after reading what was written a few posts above, when the elevator goes up, acceleration is going up... in order to find the mass of the person, what you use sum of all forces acting on elevator = ma?
so the force of gravity when it starts and the force opposing, the normal force? How do you begin to to solve part (a) in the question, the weight of the person? ... n=mg? weight is the opposition of n?... would weight be negative value then since it opposing normal force/

10. Oct 22, 2005

### Diane_

Think of it from the point of view of a scale the person is standing on. If the elevator is at rest (or at any constant velocity, for that matter), the scale is only going to "feel" the weight of the person on it - it'll read normally. If the elevator is accelerating upwards, though, the scale will be "pushing up" on the soles of the person's feet. In addition to his weight, the scale will also feel the reaction to that force - the weight and the normal force will add to produce a larger reading.

Can you see how this would work in reverse if the elevator were accelerating downward? It might help to visualize what it would look like to the scale if the elevator were to enter free fall.

11. Oct 22, 2005

### Staff: Mentor

Consider the sum of all forces acting on the person.
There are two forces acting on the person: (1) his weight, which is the pull of gravity on him, and (2) the force that the elevator exerts on him, called the normal force. You are given that normal force, which is also called the "apparent" weight, since, as Diane_ explained, that's the force that you feel.
You will apply Newton's 2nd law for each case. The weight (mg) acts down, but the normal force acts up. If you choose up to be positive, then the weight is negative and the normal force is positive. Realize that the acceleration also has a direction and sign.

12. Oct 23, 2005

### physixnot4me

applying newtons 2nd law

for question #3) above with the elevator, every says to apply newtons second law for BOTH cases ( i dont understand what the case(s) are)

is it: 591 = m(g+a) and 391= m(g-a) ???
is that how your suppose to find the weight? I really dont understand what i'm trying to be solving for.

as for question #2) above, for the microwave question, when solving for the angle why is it theta=arctan centripal acceleration/gravity? I've figured out that much, i just dont comprehend the theory behind it.

13. Oct 24, 2005

### Staff: Mentor

Yes. One case is when the elevator starts, the other when it stops.
The two unknowns are mass and acceleration. Solve for them. (Weight is just mg.)