Whats the force applied for a cart traveling up a hill?

In summary, the conversation discusses finding the power developed by a cart traveling up a hill at a constant velocity. The expert suggests using the formula P=Fv and prompts the individual to consider the forces acting on the cart, including gravity and applied force. They also recommend drawing a free-body diagram and taking components of the forces along the incline to find the relation between mg and Fa. Ultimately, the expert helps the individual understand the concept and arrive at a solution.
  • #1
AndroidX7
17
0

Homework Statement



A cart travels up a hill at a constant velocity in 2.5 (s). What power is developed by the cart?
m=120 kg
meters=12m
θ=21°

Homework Equations



P=w/t
w=F(parallel)×meters×cosθ

The Attempt at a Solution


P= w/t= Fdcosθ/2.5= F(12)(.93)/2.5= 11,2F/2.5= 4.5F
How do I find F?
 
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  • #2
Tip : Using the formula P=FIIv would be an easier approach(Why?)

The why also tells you the method of finding F.

Hint : What forces act on the car?
What is an object's acceleration when it travels with uniform velocity?

Hope this helps,
Qwertywerty.
 
  • #3
Qwertywerty said:
Tip : Using the formula P=FIIv would be an easier approach(Why?)

The why also tells you the method of finding F.

Hint : What forces act on the car?
What is an object's acceleration when it travels with uniform velocity?

Hope this helps,
Qwertywerty.
The problem I'm thinking is that F= ma and so without acceleration, how is there Force?
Is something just flying past me here?
 
  • #4
Wait. I just got it.
F=0 therefore P=0.
I think I just wasn't expecting a trick question.
 
  • #5
AndroidX7 said:
The problem I'm thinking is that F= ma and so without acceleration, how is there Force?
Is something just flying past me here?
We typically say Fnet=ma. Thus, the correct conclusion to be drawn is..?
 
  • #6
AndroidX7 said:
Wait. I just got it.
F=0 therefore P=0.
I think I just wasn't expecting a trick question.
I'm afraid that's wrong;)
 
  • #7
Qwertywerty said:
I'm afraid that's wrong;)
If I'm not mistaken "net" means Σ correct?
 
  • #8
Yes. So what forces, or components of forces, are acting on the car, along the incline?
 
  • #9
Qwertywerty said:
Yes. So what forces, or components of forces, are acting on the car, along the incline?
F normal
F gravity
F applied
 
  • #10
AndroidX7 said:
F normal
F gravity
F applied
Out of these, which forces affect the car along the incline?

Hint : If you can figure out F applied somehow, you would be able to find the power developed using the formula P=F.v
 
  • #11
Qwertywerty said:
Out of these, which forces affect the car along the incline?

Hint : If you can figure out F applied somehow, you would be able to find the power developed using the formula P=F.v
Well I know that n and g (although now that I think about it it might be g and θ) combine to find the minimal force you need to overcome gravity and inversely shows Fa, but I don't think I know how that'd work.
 
  • #12
Hmm..I'm going to try squeezing in a bit more. Do you know how to draw a free-body diagram? If you do, draw one for the car.

Now, draw the components along the incline. The net force along the incline must be zero, as we have seen earlier. Consider mg to be the force of gravity and a random variable , say F, for force exerted by the car. You will see that that F is independent of the normal.

Can you find out the relation between mg and F, using all that has been mentioned so far?
 
  • #13
I'd also reccomend you reading more theory related to this stuff.
 
  • #14
Qwertywerty said:
Hmm..I'm going to try squeezing in a bit more. Do you know how to draw a free-body diagram? If you do, draw one for the car.
say F, for force exerted by the car. You will see that that F is independent of the normal.

Can you find out the relation between mg and F, using all that has been mentioned so far?
By this I'm assuming that F= mg,
but according to my notes, as far as I can gather, F normal must equal Fa due to gravity, but your saying it's independent.
 
  • #15
Qwertywerty said:
I'd also reccomend you reading more theory related to this stuff.
where?
 
  • #16
AndroidX7 said:
By this I'm assuming that F= mg,
but according to my notes, as far as I can gather, F normal must equal Fa due to gravity, but your saying it's independent.
As I said at the start of my post, you will need to make use of a free-body diagram - your assumption is, unfortunately, wrong.

Also, I said force that the car exerts is independent of the normal. Normal force is related to gravity, but that is not relevant here. As I said earlier, you need to construct a diagram to understand what's going on.
 
  • #17
AndroidX7 said:
where?
I will not be able to help you there. You should post this question in the Science Education's textbooks forum, or something similar.
 
  • #18
Edit.
 
  • #19
Qwertywerty said:
As I said at the start of my post, you will need to make use of a free-body diagram - your assumption is, unfortunately, wrong.

Also, I said force that the car exerts is independent of the normal. Normal force is related to gravity, but that is not relevant here. As I said earlier, you need to construct a diagram to understand what's going on.
So then there's no relevant relation between mg and Fn.
Because i was hoping cos21 may subtract from mg to acquire Fn.
Am I going about this incorrectly?
 

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  • #20
AndroidX7 said:
So then there's no relevant relation between mg and Fn.
Because i was hoping cos21 may subtract from mg to acquire Fn.
Am I going about this incorrectly?
-In this question. I hope I've cleared that satisfactorily.

You've gotten the basic diagram correct. Take components of the forces along the incline.
 
  • #21
Qwertywerty said:
-In this question. I hope I've cleared that satisfactorily.

Take components of the forces along the incline.
Do you mean to add the vectors?
Or I need to account for a (-) Force down the incline so F a has acceleration... but that would still amount to Fnet and would in turn need a.
Well I'm really getting nowhere awful fast.
 
  • #22
Qwertywerty said:
-In this question. I hope I've cleared that satisfactorily.

You've gotten the basic diagram correct. Take components of the forces along the incline.
I think I may have gotten it.
You need to use trig to make a triangle with g and Fa as sides.
While I can't find the exact force, I can find the minimum required to keep it stationary. Its what the book is looking for, but in the real world, how could I find the exact force with what little I have?
 
  • #23
AndroidX7 said:
While I can't find the exact force, I can find the minimum required to keep it stationary. Its what the book is looking for, but in the real world, how could I find the exact force with what little I have?
1. Net force zero does not imply that a body is stationary. It means net acceleration is zero; and hence, it can move with constant velocity/ be stationary.

2. The question asks you for the Fa which gives net force zero. Anymore force, and the car would have a net acceleration upwards. So the word 'minimum' force is not correct. It is a particular value of force that gives zero acceleration, or constant velocity.
Thta is what you have to find.
 
  • #24
AndroidX7 said:
I think I may have gotten it.
You need to use trig to make a triangle with g and Fa as sides.
This is a little vague.
I'll try again. All the question requires is for you to take components of forces along the incline. Do you know how to do this?
 
  • #25
Qwertywerty said:
This is a little vague.
I'll try again. All the question requires is for you to take components of forces along the incline. Do you know how to do this?
I believe this is something along the lines correct?
 

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  • #26
Could you explain to me, what you did after finding x?

P.S. Seems correct. Want to know whether you used correct reasoning.
 
  • #27
I solved for power.
 
  • #28
And which force's exerted power did you find?
 
  • #29
Fa up the incline
 

1. What is the formula for calculating the force applied for a cart traveling up a hill?

The formula for calculating the force applied for a cart traveling up a hill is F = mgh + ma, where F is the force applied, m is the mass of the cart, g is the acceleration due to gravity (9.8 m/s^2), h is the height of the hill, and a is the acceleration of the cart.

2. How does the angle of the hill affect the force applied for a cart traveling up?

The angle of the hill affects the force applied for a cart traveling up because it determines the vertical height the cart must travel. The steeper the angle, the greater the force required to overcome the force of gravity and move the cart up the hill.

3. Is the force applied for a cart traveling up a hill the same as the force applied for a cart on a flat surface?

No, the force applied for a cart traveling up a hill is not the same as the force applied for a cart on a flat surface. This is because on a flat surface, the force required to move the cart is only to overcome friction, whereas on a hill, the force must also overcome gravity.

4. How does the weight of the cart affect the force applied for traveling up a hill?

The weight of the cart directly affects the force applied for traveling up a hill. The heavier the cart, the more force is needed to overcome the force of gravity and move the cart up the hill.

5. What other factors besides mass and incline angle can affect the force applied for a cart traveling up a hill?

Other factors that can affect the force applied for a cart traveling up a hill include the friction between the wheels of the cart and the surface of the hill, any external forces such as wind or resistance, and the design and condition of the cart itself.

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