Solving Physics Problems: Forces and Acceleration

Click For Summary

Homework Help Overview

The discussion revolves around three physics problems related to forces and acceleration, specifically involving static friction, net forces, and equilibrium conditions in various scenarios. The problems involve a cup of coffee in an accelerating airplane, a water skier being pulled by a force, and a student pressing a book between their hands.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the concept of static friction and its role in determining maximum acceleration without sliding. They discuss the relationship between forces acting on the skier and the conditions for constant velocity. The role of normal force in the context of the book being pressed is also examined, with questions about how to calculate it and its relationship to static friction.

Discussion Status

Some participants have provided guidance on identifying forces and drawing free-body diagrams. There is an ongoing exploration of the relationships between forces, accelerations, and the conditions necessary for equilibrium in the various scenarios presented.

Contextual Notes

Participants note the importance of showing work and understanding the underlying physics concepts, such as static friction and net forces, to facilitate better assistance. There is an acknowledgment of the need for clarity in problem setup and assumptions.

1moretry
Messages
2
Reaction score
0
i'm trying to finish up my online homework before the due date tonite at midnite. i have the rest perfect, just these three got me stumped.

7. [CJ6 4.P.038.] A cup of coffee is sitting on a table in an airplane that is flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.20. Suddenly, the plane accelerates, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?




11. [CJ6 4.P.088.] A 75 kg water skier is being pulled by a horizontal force of 525 N and has an acceleration of 2.5 m/s2. Assuming that the total resistive force exerted on the skier by the water and the wind is constant, what force is needed to pull the skier at a constant velocity?




12. [CJ6 4.P.089.] A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs 29 N. The coefficient of static friction between his hands and the book is 0.31. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?
 
Physics news on Phys.org
General Ettiquette

It's a good idea to post questions separately. It's also generally a good idea to show some work. Otherwise people will not help you because they're not interested in doing your homework for you, and it doesn't look like you're trying.

That said, what do you know about static friction?

Can you list the forces in each situation, or write equations for them?
 
whoops, my bad. i'll keep that in mind next time. thanks for not blowing up on me. this is what i got so far.

7. i really don't know where to go with this one. all i got is...

static frictional force = (coefficient of static friction) x (normal force)

fs = (.20)(Fn)

i'm trying to find the acceleration, so i'd have to use F=ma somewhere, but i don't have a mass, and the only acceleration i know is gravity.

11. the skier is 75kg, so the force of weight would be 735N down, and so the normal force would be 735N up, since there are no other up/down forces and that he's being pulled horizontally.

the question is what force is needed to pull the skier at a constant velocity, so that would mean zero acceleration, right? or no? i don't see where i can plug 0 into an equation like F=ma.

12. i understand that it is the static friction that is keeping the book from falling between the hands. so that would mean that the force of static friction is equal to the force of the weight of the book.

static frictional force = (coefficient of static friction) x (normal force).

fs = (.31)(29N) = 8.99

what can i do with this to find out the horizontal forces? they should be equal in this case, too, right?
 
Last edited:
7. What is the normal force equal to ?

11. You are given the force that the skier is being pulled with (525N). His mass is 75 kg. If this were the only horizontal force on the skier, his acceleration would be a = F/m = 525/75 = 7 m/s^2. But his real acceleration is less than this. Why ?

12. The normal force is not the weight of the book. The normal force is the force between the surfaces that are touching each other, ie : the force between the hand and the book cover.

For all these problems, the first thing to do is draw a free-body (force) diagram. Surely your text shows you how to do this.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
13
Views
4K
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 42 ·
2
Replies
42
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
4
Views
2K
Replies
7
Views
2K