Recent content by Patta1667

Have you done any work on the problem? Try drawing it out and labeling the distances, do you know how to find the length of the sides of an isosceles (two sides same length) triangle?

Homework Statement I'm not given any signs of a correct answer in the book, so could I get a check here? A car is driven on a large revolving platform which rotates with constant angular speed w. At t = 0 a driver leaves the origin and follows a line painted radially outward on the...

Okay, thanks :smile:. Some sort of brain block there!

Then how can I know the horizontal tension in the rope in order to find the total tension at an end?

Do you mean that the total tension is W/2? I cannot see anything when it comes to ropes, but I have read in a book that the horizontal tension is constant and it makes sense that T_x = T \sin \theta . Using this equation and T_y = (W/2) \cos \theta I can find the tension at an end for a...

Homework Statement A uniform rope of weight W hangs between two trees. The ends of the rope are the same height, and each make angle \theta with the trees (the angle is between the rope and the lower part of the tree). Find the tension at either end of the rope. Homework Equations...

Sure, sure. Do you understand the method now, any related questions?

1) Could you sum it up in one post please? :) 2) Alright, I'll settle this. You need to consider the forces acting up and down the slope. If you have a cart of mass m on a slope, the component down the slope is mgsin(theta), and the other component is normal to the slope. If you push a...

Thank you! I had taken N = mgcos(theta) in all my equations, I completely forgot that the inertial force should be considered. The answer comes right out now, thanks!

Ooops, let \theta = \pi/4 to get that answer... sorry! Well, could you lead me in the right direction please on how you obtained that?

Okay, the maximum friction force along the surface is \mu N , so the horizontal component is f \cos \theta = \mu N \cos \theta = \mu m g \cos^2 \theta . This means the block has an acceleration a = \mu g \cos^2 \theta horizontally if friction alone is responsible. Unfortunately, using...

Consider the forces acting up and down the incline. Down the incline, you have a component of the force of gravity. Up the incline you have a component of the horizontal force of the pusher. Using F = ma, and F = (component of force) - (component of gravity), you find the force F. Hint...

Homework Statement A block rests on a wedge inclined at angle \theta . The coefficient of friction is \mu . The wedge is then given a horizontal acceleration a. Assuming that \tan (\theta) < \mu , find the minimum acceleration for the block to remain on the wedge without sliding...

I'm heading into my senior year of high, but I'll be taking some sort of general physics class at a local uni in the fall so I need to prepare bigtime! I'll make sure to write down Morin's name, thanks!

Yup, I figured K & K looked like a good intro to mechanics choice. Did you encounter this book yourself or from classes? Thanks! I see the mistake now, it works: A = x'' + y'' mx'' = \frac{N}{\sqrt{2}} my'' = \frac{N}{\sqrt{2}} - mg This means: x'' = y'' + g \implies A...