Shopping Cart Force Calculation

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This should give you Force1, or what the person exerts, in terms of known quantities.In summary, a person pushing a 16.0-kg shopping cart at a constant velocity for a distance of 22.0 m in a direction 29° below the horizontal is faced with a 48.0 N frictional force opposing the motion. The magnitude of the force that the shopper exerts can be found by using Newton's Second Law and setting the sum of the forces in the horizontal direction to zero. This results in the force exerted by the person being equal to the horizontal component of the pushing force, which can be calculated using trigonometry. The answer is 54.9 N.
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Homework Statement


A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes in a direction 29° below the horizontal. A 48.0 N frictional force opposes the motion of the cart. What is the magnitude of the force that the shopper exerts?

Homework Equations


F = mg
W = (F cosθ)s

The Attempt at a Solution


1. F = (16kg)(9.8m/s2) = 156.8 N
2. FN= 156.8 N - 48.0 N = 108.8 N

Correct answer: 54.9 N

I know if you divide 108.8 by two you get 54.4 (which is really close to the answer), but I don't think you're supposed to divide by 2. Can someone help me out? :s
 
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  • #2
You're told that the cart moves at a constant velocity, i.e. there's no NET force acting on the cart. In other words, the sum of the forces in the horizontal direction is zero. I would suggest drawing a free body diagram if you haven't already. The mass of the cart and the distance it moves is irrelevant here. Use Newton's Second Law.
 
  • #3
Cursed, in your solution you subracted a vertical force, the weight of the mass, by the friction, a horizontal. This is invalid. The weight is in the vertical direction, and the force of friction always points in the direction opposite of the motion, which is in this case horizontal. To subract or add forces, they must have the same direction.
 
  • #4
Thanks, Markus.

I'm still not getting the answer, though. (I used F = ma, and I had the force be 156.8 N.)

1. tan(29°) = 156.8 N/x; x = 282.8746 N
2. cos(29°) = z/48; z = 41.9817 N
3. 282.8746 - 41.9817 = 240.8929

What am I doing wrong? :/
 
  • #5
How would you set up Newton's Second Law for the horizontal direction? You'll end up with a very simple equation.
 
  • #6
Cursed,

The weight of the mass is completely irrelevant for this question. The mass is on some horizontal surface, and by Newton's Third Law, it is canceled out by a normal force.

The unknown here is what force the person exerts. We'll call this Fext. Only the horizontal component of this force contributes to motion. The vertical does not. Try pushing down on a brick and see if it moves! So using trig, the horizontal component of the force exerted is Fext x cos29. (DO NOT take my word for it, come to this conclusion yourself!)

Since it is stated in the problem that the velocity is constant, then there is no acceleration. If there is no acceleration there is no net force on the mass. So, we can equation something to zero here... After you've figured out what, the answer is easily obtained.
 
  • #7
MarkusNaslund19 said:
Cursed,

The weight of the mass is completely irrelevant for this question. The mass is on some horizontal surface, and by Newton's Third Law, it is canceled out by a normal force.

The unknown here is what force the person exerts. We'll call this Fext. Only the horizontal component of this force contributes to motion. The vertical does not. Try pushing down on a brick and see if it moves! So using trig, the horizontal component of the force exerted is Fext x cos29. (DO NOT take my word for it, come to this conclusion yourself!)

No matter what I do, I can't get the answer:

1. cos29 * F = (F * cos29)s
2. (cos29 * ma)s = mg
3. (cos29 * F)s = 0
4. (ma * cos29)s = 0.5mv2

MarkusNaslund19 said:
Since it is stated in the problem that the velocity is constant, then there is no acceleration. If there is no acceleration there is no net force on the mass. So, we can equation something to zero here... After you've figured out what, the answer is easily obtained.

What am I supposed to set equal to zero?
 
  • #8
Cursed said:
No matter what I do, I can't get the answer:

1. cos29 * F = (F * cos29)s
2. (cos29 * ma)s = mg
3. (cos29 * F)s = 0
4. (ma * cos29)s = 0.5mv2



What am I supposed to set equal to zero?

You need to read the above posts more carefully. You've been told that the mass of the object and distance it travels are of no consequence, so don't use mass or distance.. at all. The "key" to solving this is Newton's Second Law. Look at the free body diagram below:

http://s62.photobucket.com/albums/h116/pepsi_in_a_can/?action=view&current=force.jpg

What two forces are acting in the x-direction? More specifically, what component of the pushing force are we concerned with? And what force opposes it such that the cart moves at a constant velocity? Call the opposing force Force2, and the component it works against Force1.

Once you know this, you can use Newton's Second law [tex] \Sigma F_x = ma_x [/tex] to sum the forces in the x-direction:

[tex] \Sigma F_x = Force1 + Force2 = ma_x [/tex]

But what should this equation equal seeing as the velocity is constant?
 
Last edited:

What is the "Shopping Cart Problem"?

The "Shopping Cart Problem" is a thought experiment that explores the ethical dilemma of whether or not it is permissible to take an item from someone else's shopping cart if they have left it unattended.

Why is the "Shopping Cart Problem" important to study?

The "Shopping Cart Problem" raises important questions about morality, personal responsibility, and the social contract. It also reveals the complexities of real-life situations and decision-making processes.

What are the main arguments for taking an item from someone else's shopping cart?

One argument is that if the person has left their cart unattended, they have implicitly given up their right to the items inside. Another argument is that taking an item from an unattended cart is not significantly different from finding a lost item on the ground.

What are the main arguments against taking an item from someone else's shopping cart?

One argument is that it is morally wrong to take something that does not belong to you, regardless of the circumstances. Another argument is that it goes against the social norm of respecting others' property and could lead to a breakdown of trust in society.

Is there a clear answer to the "Shopping Cart Problem"?

There is no universally agreed-upon answer to the "Shopping Cart Problem." It ultimately depends on an individual's personal values and beliefs. Some may argue that it is never okay to take something from another person's cart, while others may believe it is acceptable under certain circumstances.

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