Newton's second law problem with friction

In summary: If so, you can annotate them with information about the forces and how they change when the worker pulls upward.In summary, the frictional force opposing the horizontal push is reduced when the worker pulls upward.
  • #1
isukatphysics69
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1. Homework Statement

i did part a,b,c but i am confused about part d.. i have a test tomorrow and will reattempt this when i wake up but i really need a hint here..so when the worker pulls upward he is decreasing the magnitude of -mg right? so if Fn and mg initially cancel each other out, when the worker pulls upward he is making it so that the acceleration in the y direction is no longer 0 correct? so netforceydirection = Fn-mg+Fpull and it is non zero?

Homework Equations

The Attempt at a Solution

 

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  • #2
isukatphysics69 said:
i did part a,b,c but i am confused about part d.. i have a test tomorrow and will reattempt this when i wake up but i really need a hint here..so when the worker pulls upward he is decreasing the magnitude of -mg right? so if Fn and mg initially cancel each other out, when the worker pulls upward he is making it so that the acceleration in the y direction is no longer 0 correct? so netforceydirection = Fn-mg+Fpull and it is non zero?

the frictional force opposing the horizontal push is = coefficient of friction x R (the net reaction)
initially R= weight mg of the crate
after help say of F1 upward F1+ R = mg

so R= mg- F1 thereby reducing the frictional force = mux(mg-F1)
in the horizontal direction. so it will help him push easily
 
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  • #3
You really need to use the homework template more effectively. What you've put in the problem statement belongs in the Attempt at solution section, and you should list any equations that you think are relevant to this type of problem in the Relevant equations section. See the pinned thread, Guidelines for students and helpers, for tips on how to create effective homework help requests.

An empty Attempt at solution section is a major flag to moderators to scrutinize your post to judge whether or not you've put any effort so far. Helpers, who are volunteering their expertise, shouldn't have to tease out your efforts from random places in the post.

Considering the points you've raised in your post, I am wondering whether or not you've drawn Free Body Diagrams for the scenarios. Have you done that?
 
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1. What is Newton's second law?

Newton's second law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. This can be represented mathematically as F=ma.

2. How does friction affect Newton's second law?

Friction is a force that opposes motion, so it acts in the opposite direction of the applied force. This means that when considering Newton's second law, the frictional force must be taken into account and subtracted from the applied force.

3. How do you solve a problem involving Newton's second law with friction?

To solve for the acceleration of an object in a problem involving Newton's second law with friction, you must first identify all the forces acting on the object and their respective directions. Then, use the formula F=ma to calculate the net force and solve for the acceleration. Finally, take into account the frictional force and adjust the net force accordingly.

4. What are some common examples of Newton's second law problems with friction?

Some common examples of Newton's second law problems with friction include a car accelerating on a road, a person pushing a heavy object across the floor, or a hockey puck sliding on ice. In all of these scenarios, the frictional force acts in the opposite direction of the applied force and must be taken into account when solving for the acceleration.

5. How does the coefficient of friction affect Newton's second law problem with friction?

The coefficient of friction is a measure of how much friction is present between two surfaces. It can affect Newton's second law problem with friction by either increasing or decreasing the frictional force acting on an object. A higher coefficient of friction means a stronger frictional force, while a lower coefficient of friction means a weaker frictional force.

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