Understanding Forces and Work: A) to C)

In summary, the conversation discusses the relationship between work, force, and displacement in a scenario where a shopper pushes a cart at different angles. It is determined that the work remains the same when the force is smaller, as long as the displacement is in the same direction as the force. The conversation also discusses the horizontal and vertical components of forces and how they affect the overall magnitude of the force.
  • #1
omarMihilmy
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ImageUploadedByPhysics Forums1387228165.386829.jpg


Okay so in number a) I got the work just fine
In number b) pushing horizontally meaning that cos angle is 0 which would give a larger force?
The correct answer is smaller
In c) the work done on the cart by the shopper how would it be the same if the force is smaller let us say?
 
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  • #2
pushing the cart at a 0° angle would not produce a greater force--it would increase the work, if you calculate it with the original displacement and applied force: W = (50m)(35N)cos(0°). but the question is asking how the applied force would have to change, if at all, for the speed to stay the same? you can assume the displacement is the same, since it says she goes down "the next aisle."
 
  • #3
So you are saying for the speed to stay the the same the force will be smaller ? Is that correct?
What about the pushing horizontally part won't it affect at all the force?

Secondly in c) the work will remain the same how is it possible if the force is smaller ?

Please Help!
 
  • #4
omarMihilmy said:
So you are saying for the speed to stay the the same the force will be smaller ?
The speed doesn't matter. As long as the speed is constant, the forces are in balance.
In each scenario, there are essentially four forces on the cart, right? The gravitational force, Mg, the shopper's applied force Fs, the frictional force Ff, and the normal force N from the ground.
Which direction does Ff act in?
What is the horizontal component of Fs equal to?
What is the vertical component of Fs equal to?
 
  • #5
haruspex said:
The speed doesn't matter. As long as the speed is constant, the forces are in balance.

In each scenario, there are essentially four forces on the cart, right? The gravitational force, Mg, the shopper's applied force Fs, the frictional force Ff, and the normal force N from the ground.

Which direction does Ff act in?

What is the horizontal component of Fs equal to?

What is the vertical component of Fs equal to?

In b)

The friction force acts opposite to the direction of motion

Vertical component will be equal to mg

Horizontal will be equal to Fs
 
  • #6
omarMihilmy said:
In b)
Since we're trying to compare (a) with (b), best is to answer for an unknown angle of applied force, downwards at angle θ say, then see how the magnitude of Fs depends on θ.
The friction force acts opposite to the direction of motion
So horizontal, yes.
Vertical component will be equal to mg
The vertical component of the shopper's push will equal mg?? The shopper is pushing the cart along the ground, not carrying it.
Horizontal will be equal to Fs
When pushing horizontally, yes, but I meant in terms of the frictional force.
 
  • #7
Then the frictional force is equal to his horizontal force true. Please see attachment and tell me if this is correct

ImageUploadedByPhysics Forums1387276263.270333.jpg

Is this where it comes to say that the force is smaller?
ImageUploadedByPhysics Forums1387276295.453962.jpg
 
  • #8
omarMihilmy said:
Then the frictional force is equal to his horizontal force true. Please see attachment and tell me if this is correct

View attachment 64873
Is this where it comes to say that the force is smaller?
View attachment 64874

Yes.
Now what about the work? Remember work done ( by a constant force) = force x what?
 
  • #9
Work = Force x Displacement
Force is smaller now as we agreed
And the displacement is the same

ImageUploadedByPhysics Forums1387278129.579551.jpg
 
  • #10
This is not correct the work should be the same whyyyyyy?
 
  • #11
omarMihilmy said:
Work = Force x Displacement
But what displacement, exactly? Think about the directions.
 
  • #12
It is displaced by 50 meters in the next aisle right?
 
  • #13
omarMihilmy said:
It is displaced by 50 meters in the next aisle right?
No, I mean in the statement "work = force x distance", it isn't just any old distance. The distance has a specific relationship to the force. It's to do with the direction of each.
 
  • #14
Okay I see that please elaborate more how will it be smaller?
 
  • #15
omarMihilmy said:
Okay I see that please elaborate more how will it be smaller?
It's the distance in the direction of the force. In vector terms, and allowing for a variable force, it's ∫F.ds, i.e. the integral of the dot product of the force vector with the distance element vector. For a constant force you can simplify that to F.s, again that's the dot product of vectors. If the angle between the force vector and the distance vector is θ then you can write it as |F||s|cos(θ).
So, it doesn't matter whether you consider the distance component in the direction of the force or the force component in the direction of the distance moved - it comes to the same thing.
In the present problem, the movement is horizontal. So you need to compare the horizontal components of the forces for the two cases.
 
  • #16
haruspex said:
It's the distance in the direction of the force. In vector terms, and allowing for a variable force, it's ∫F.ds, i.e. the integral of the dot product of the force vector with the distance element vector. For a constant force you can simplify that to F.s, again that's the dot product of vectors. If the angle between the force vector and the distance vector is θ then you can write it as |F||s|cos(θ).

So, it doesn't matter whether you consider the distance component in the direction of the force or the force component in the direction of the distance moved - it comes to the same thing.

In the present problem, the movement is horizontal. So you need to compare the horizontal components of the forces for the two cases.
But the magnitude of the force is smaller?
 
  • #17
omarMihilmy said:
But the magnitude of the force is smaller?
The overall magnitude, yes, but what about the horizontal components of each? Are they different?
 
  • #18
Ohhh yes I got it finally the the force(a) horizontal component = force (b)

Thank you very much !
 

FAQ: Understanding Forces and Work: A) to C)

1. What are forces and how do they affect motion?

Forces are physical interactions between objects that can cause a change in their motion. They can push, pull, or twist an object and can either speed it up, slow it down, or change its direction.

2. How is the concept of work related to forces?

Work is the result of a force acting on an object to cause a displacement. It is calculated by multiplying the magnitude of the force by the distance the object moves in the direction of the force.

3. What is the difference between scalar and vector quantities in relation to forces?

Scalar quantities only have a magnitude or size, while vector quantities have both magnitude and direction. Forces are vector quantities, as they have both magnitude and direction.

4. How do we measure forces and work?

Forces are measured in Newtons (N) using a device called a spring scale. Work is measured in joules (J) and can be calculated using the equation work = force x distance.

5. Can forces and work be negative?

Yes, forces and work can be negative. A negative force indicates that it is acting in the opposite direction of the reference direction, while negative work occurs when the force and displacement are in opposite directions.

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