Understanding Forces and Work: A) to C)

[omarMihilmy]

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?

 

[shanname]

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."

 

[omarMihilmy]

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!

 

[haruspex]

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 F_s, the frictional force F_f, and the normal force N from the ground.
Which direction does F_f act in?
What is the horizontal component of F_s equal to?
What is the vertical component of F_s equal to?

 

[omarMihilmy]

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 F_s, the frictional force F_f, and the normal force N from the ground.

Which direction does F_f act in?

What is the horizontal component of F_s equal to?

What is the vertical component of F_s 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 F_s

 

[haruspex]

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 F_s 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 F_s

When pushing horizontally, yes, but I meant in terms of the frictional force.

 

[omarMihilmy]

Then the frictional force is equal to his horizontal force true. Please see attachment and tell me if this is correct

Is this where it comes to say that the force is smaller?

 

[haruspex]

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?

 

[omarMihilmy]

Work = Force x Displacement
Force is smaller now as we agreed
And the displacement is the same

 

[omarMihilmy]

This is not correct the work should be the same whyyyyyy?

 

[haruspex]

Work = Force x Displacement

But what displacement, exactly? Think about the directions.

 

[omarMihilmy]

It is displaced by 50 meters in the next aisle right?

 

[haruspex]

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.

 

[omarMihilmy]

Okay I see that please elaborate more how will it be smaller?

 

[haruspex]

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.

 

[omarMihilmy]

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?

 

[haruspex]

But the magnitude of the force is smaller?

The overall magnitude, yes, but what about the horizontal components of each? Are they different?

 

[omarMihilmy]

Ohhh yes I got it finally the the force(a) horizontal component = force (b)

Thank you very much !