Calculating Forces in a Moving Lift

  • Thread starter Thread starter SuperNova1
  • Start date Start date
  • Tags Tags
    Forces Lift
Click For Summary

Homework Help Overview

The problem involves calculating forces acting on a block in a moving lift, specifically focusing on scenarios where the lift accelerates upwards, moves at constant speed, and decelerates upwards. The subject area includes dynamics and forces, particularly the concepts of normal force and gravitational force.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of forces acting on the block, including the normal force and gravitational force. There are attempts to clarify the definitions of net force and resultant acceleration, as well as the implications of different scenarios (acceleration, constant speed, deceleration).

Discussion Status

Some participants have provided guidance on identifying the normal force and its relationship to gravitational force. There is an ongoing exploration of the correct interpretation of forces and accelerations, with multiple interpretations being discussed. Participants are questioning assumptions and clarifying concepts without reaching a consensus.

Contextual Notes

Participants note discrepancies in the understanding of gravitational acceleration and its application in calculations. There is also a recognition of imposed homework rules that may limit the scope of discussion.

SuperNova1
Messages
10
Reaction score
0

Homework Statement



(in these questions we were told that g+ 10ms/s i know its not strictly true g should be 9.81 but our teacher told us to do this)

2. A 35 kg block rests on the floor of a lift.
Find the force between the floor and the block if the lift (i) accelerates upwards at 4 ms-2.
(ii) travels upwards at constant speed
(iii) decelerates upwards as it comes to a stop at 7 ms-2.


Homework Equations



F(net)=m*a


The Attempt at a Solution



(i) Accelerates upwards at 4 ms/s

Mass = 35kg
g(force due to gravity)= 10 m/s/s
Resultant acceleration = 10 + 4 = 14m/s/s
Force(net) = mass x acceleration
Force (net) = 35 x 14
Force (net) = 490N


(ii) Constant speed

Mass = 35kg
g(force due to gravity) = 10m/s/s
Resultant acceleration = 10 – 0(constant speed = 0 acceleration)
Force(net) = mass x acceleration
Force (net) = 35 x 10
Force(net) = 350N

(iii) Decelerates upwards as it comes to a stop at 7m/s/s

Mass = 35
g(force due to gravity) = 10m/s/s
acceleration upwards = -7m/s/s
resultant acceleration = 10 – 7 = 3m/s/s
Force(net) = mass x acceleration
Force(net) = 35 x 3
Force(net) = 105N
 
Physics news on Phys.org


If it is said that an object accelerates with 4 m/s2 upward it means that the "resultant" acceleration is 4 m/s2.

The acceleration is the result of the forces acting on the object: ma = ∑Fi. What forces act on a body lying on a horizontal surface?

ehild
 


ehild said:
If it is said that an object accelerates with 4 m/s2 upward it means that the "resultant" acceleration is 4 m/s2.

The acceleration is the result of the forces acting on the object: ma = ∑Fi. What forces act on a body lying on a horizontal surface?

ehild
thank you for your reply

In what way would i calculate the force between the lift and the mass? would i just use 4 as the acceleration and go F = M * a
F= 35 * 4
F= 140N
would this be correct?
 


This is the resultant of the forces acting on the block. You need to find the force the floor exerts on the block, that is, the normal force, N.
So there are two forces, the normal force N and gravity G=mg.

Draw the forces and the acceleration. What is the direction of the normal force with respect to G? what is the direction of the acceleration with respect to G?

ehild
 


ehild said:
This is the resultant of the forces acting on the block. You need to find the force the floor exerts on the block, that is, the normal force, N.
So there are two forces, the normal force N and gravity G=mg.

Draw the forces and the acceleration. What is the direction of the normal force with respect to G? what is the direction of the acceleration with respect to G?

ehild

But then all i would be doing is adding on the weight because the weight (M*g) is acting vertically downwards and the lift is accelerating vertically upwards which means the force must be acting upwards. the net force is surely 140 + the weight because these forces are acting against each other which means that the total force will be greater than either of the two forces. so i calculate weight (35*10) which gives 350 add that to 140 and i get 490N my original answer, what's wrong here? (apart from the acceleration due to gravity)
 


SuperNova1 said:
But then all i would be doing is adding on the weight because the weight (M*g) is acting vertically downwards and the lift is accelerating vertically upwards which means the force must be acting upwards.
You mean the net force, N-mg which is equal to ma=35*4=140 N when the lift accelerates upward. Indicate clearly which force you speak about.

SuperNova1 said:
the net force is surely 140 + the weight because these forces are acting against each other which means that the total force will be greater than either of the two forces. so i calculate weight (35*10) which gives 350 add that to 140 and i get 490N my original answer, what's wrong here? (apart from the acceleration due to gravity)

The number is correct, but it is not the net force.

In case of constant velocity, the acceleration is zero. It is wrong to say that

"g(force due to gravity) = 10m/s/s": g is acceleration, not force.

"Resultant acceleration = 10 – 0(constant speed = 0 acceleration)"

This is totally wrong. The acceleration is not 10-0=10, but 0.

"Force(net) = 350N" Wrong again. The net force is zero, as the acceleration is zero:N-mg=0. The normal force is 350 N, that would be the correct answer.

(iii) Decelerates upwards as it comes to a stop at 7m/s/s

"acceleration upwards = -7m/s/s
resultant acceleration = 10 – 7 = 3m/s/s"
Wrong again. The "resultant" acceleration is -7 m/s2.

"Force(net) = mass x acceleration"
"Force(net) = 35 x 3"
wrong. The net force is acceleration times mass Fnet=-35*7=-245 N
The net force is equal to gravity + normal force from the floor N (it is upward): Fnet = -mg+N=ma
-350+N=-245 =>N=-245+350=105 N.

So your numbers are correct, but your answers and explanations are wrong.
 


ehild said:
You mean the net force, N-mg which is equal to ma=35*4=140 N when the lift accelerates upward. Indicate clearly which force you speak about.



The number is correct, but it is not the net force.

In case of constant velocity, the acceleration is zero. It is wrong to say that

"g(force due to gravity) = 10m/s/s": g is acceleration, not force.

"Resultant acceleration = 10 – 0(constant speed = 0 acceleration)"

This is totally wrong. The acceleration is not 10-0=10, but 0.

"Force(net) = 350N" Wrong again. The net force is zero, as the acceleration is zero:N-mg=0. The normal force is 350 N, that would be the correct answer.

(iii) Decelerates upwards as it comes to a stop at 7m/s/s

"acceleration upwards = -7m/s/s
resultant acceleration = 10 – 7 = 3m/s/s"
Wrong again. The "resultant" acceleration is -7 m/s2.

"Force(net) = mass x acceleration"
"Force(net) = 35 x 3"
wrong. The net force is acceleration times mass Fnet=-35*7=-245 N
The net force is equal to gravity + normal force from the floor N (it is upward): Fnet = -mg+N=ma
-350+N=-245 =>N=-245+350=105 N.

So your numbers are correct, but your answers and explanations are wrong.

ah i see now, that makes sense :D thank you very much for your help, i just got to get the basis for my numbers right, i was looking at the questions in the wrong way but now i see thanks again
 

Similar threads

Replies
6
Views
3K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
23
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
3K
Replies
17
Views
2K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K