Net acceleration, free body diagrams, velocity, force normal

In summary, the student attempted to solve homework equations, but had difficulty with the directions of the forces and the algebra.
  • #1
alexandria
169
2

Homework Statement


upload_2016-3-29_20-22-38.png


Homework Equations


relevant equations are provided with each question below

The Attempt at a Solution


please check my solutions to ensure they are correct. thanks.
a) [/B]
upload_2016-3-29_20-21-57.png

b)

Calculate net force:

Fnet = FA + Ff + Fg

Fnet = (1.20 x 104 N [up]) + (1.40 x 103 N [down]) + (10094 N [down])

Fnet = (1.20 x 104 N [up]) – (1.40 x 103 N [up] + 10094 N [up])

Fnet = 506 N [up]

Calculate net acceleration:

Fnet = m a

a.net = Fnet /m

= 506 N [up] / 1030 kg

= 0.49 m/s2 [up]

c)
upload_2016-3-29_20-22-12.png
d)

Relevant Equation to solve for the force normal:

FN = mg + ma

FN = (35 kg) (9.8 N.kg [down]) + (35 kg) (0.49 m/s2 [up])

FN = 360.15 N [up]

The force normal acting on this passenger is approximately 360 N [up].

e)

Given information:

v1 = 0

a = 0.49 m/s2 [up]

t = 12.0 s

Unknown:

v2 =?

Equation:

v2 = v1 + a x total time

= 0 + (0.49 m/s2) x (12.0 s)

= 5.88 m/s [up]

The velocity is approximately 5.9 m/s [up]
 

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  • #2
You always do nice diagrams... you mostly did good, I'll add some notes for tweaks and questions to help you think about things.

(a) that's not a fbd ... just draw a box, forces should be shown coming from their point of action.
Where two or more arrows must overlap, you can "explode" the diagram, say, using a dotted line to indicate where the forces act.

(b)You were careful to indicate direction for forces... good.
A shortcut is to write which direction is positive... ie "+ve = upwards" ... then you can just convert.
The 10^4 didnt come out.

It is best practise to do the algebra first... ie. F-W-f=(W/g)a : where W is weight, F is the applied force, f is friction, and g is the acceleration die to gravity. Then a = g(F-W-f)/W ... plug numbers in as magnitudes since direction is already in the realtion.

(c) you may have too many forces... the fbd should isolate the mass in question, so do not draw the elevator in. What are the forces just on the passenger?
Note: the passenger accelerates at the same rate as the whole elevator, so you can check your answer against what you got before.

(d) this equation should come from a free body diagram

(e) works fine... you can check by sketching the v-t graph.

Well done.
 
  • #3
for c) are you saying the diagram is incorrect, regarding the force normal, gravity force, and force applied. the person accelerates at the same rate as the elevator (Fapplied), there is also gravity acting downwards on the person, and then the normal force that supports the weight of the person. You mentioned that i have too many forces, what does this imply?
 
  • #4
By your numbers: a = (FA+FN+Fg)/m ... so what is this value using:
FA=12000N, FN=360.15N, Fg= -343N, and m=35kg (your figures)?
Compare this value with that obtained for the entire elevator in part (a)

Consider: Where on the person does the "force applied" act?
 
Last edited:
  • #5
are you referring to the direction that the force applied acts on the person. So wouldn't that be up, because the elevator is going up.
 
  • #6
so for c) is this how the fbd goes:
upload_2016-3-31_14-30-30.png
i used the same forces, but just took out the elevator to isolate the passenger ?
 
  • #7
Please answer guiding questions - vis:
By your numbers: a = (FA+FN+Fg)/m ... so what is this value using:
FA=12000N, FN=360.15N, Fg= -343N, and m=35kg (your figures)?
Compare this value with that obtained for the entire elevator in part (a)
are you referring to the direction that the force applied acts on the person. So wouldn't that be up, because the elevator is going up.
No... "up" would be a direction. I have asked about a location.
eg. If I pull on your hand, the location where the force of me pulling on you is acting is "your hand".

For the lift: the applied force comes via a cable attached to the roof - so the location is "on the roof".
However - there is no cable attached to the person.

I used the same forces, but just took out the elevator to isolate the passenger ?
Isolating the passenger is good - but you should draw the arrows as forces that are attached to the passenger ... i.e. if the passenger had a rope pulling on each arm, you'd show an arrow coming from each arm. The person's weight acts from their centre of mass, for eg. and points down.
There must be an upwards force so where does it act? Does it pull the passenger up by the hair for example?
 

Related to Net acceleration, free body diagrams, velocity, force normal

1. What is net acceleration?

Net acceleration refers to the overall change in velocity of an object, taking into account all of the forces acting upon it. It is a vector quantity, meaning it has both magnitude and direction. It is calculated by dividing the net force acting on an object by its mass.

2. How do free body diagrams help in understanding motion?

Free body diagrams are visual representations of all the forces acting on an object. They help in understanding motion by showing the direction and relative magnitude of each force, and how they interact with each other to affect the motion of the object.

3. What is the relationship between velocity and acceleration?

Velocity is the rate of change of an object's position, while acceleration is the rate of change of an object's velocity. In other words, acceleration is the change in velocity over time. The two are directly related, meaning that any change in velocity will result in a corresponding change in acceleration.

4. What is force normal?

Force normal, also known as normal force, is the component of force that is perpendicular to the surface an object is resting on. It is a reaction force that prevents the object from falling through the surface. In a free body diagram, it is typically represented by a vertical arrow pointing upwards.

5. How do you calculate the net force on an object?

The net force on an object is the sum of all the individual forces acting on it. If the object is at rest or moving at a constant velocity, the net force is equal to zero. If the object is accelerating, the net force can be calculated by using the formula F=ma, where F is the net force, m is the mass of the object, and a is the acceleration.

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