forces

13
Forces The unedited story

Upload: chipo

Post on 07-Jan-2016

35 views

Category:

Documents


0 download

DESCRIPTION

Forces. The unedited story. Remember this ? The block sits on the table. What forces are there ?. R. The normal reaction. The weight. m is the mass of the block. mg. g is gravity (9.8 ms -2 ). And then Nico comes along and tips the rough table! Bad Nico - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Forces

Forces

The unedited story

Page 2: Forces

Remember this ?

The block sits on the table. What forces are there ?

The normal reaction

The weight

R

mgm is the mass of the block

g is gravity (9.8 ms-2)

Page 3: Forces

And then Nico comes along and tips the rough table! Bad Nico

Where are the forces acting now ? (The block doesn’t move)

F≤R

R

mg

m is the mass of the block

g is gravity (9.8 ms-2)

Page 4: Forces

F≤R

R

mg

Now you look at the forces acting parallel and perpendicular to the plane (the table)

Page 5: Forces

R

mgcos

So perpendicular to the table we have . . . .

R = mgcos

Page 6: Forces

mgsin

Parallel to the table we have . . . .

F≤R

If the block is at the point of moving F = mgsin

R = mgsin

Page 7: Forces

mgsin

F≤R

Now add a rope pulling the block up the slope

TCall this force T for the tension in the rope

So what happens now?

The force acting up the slope is T

The force acting down the slope is F + mgsinFRICTION HAS CHANGED DIRECTION

Page 8: Forces

mgsin

F≤R

T

So what happens when the block starts to move?

The force acting up the slope is T

The force acting down the slope is F + mgsin

F = ma

Resolving the forces gives

T – F - mgsin = ma

Page 9: Forces

And now with numbers !

A block of mass 3kg is placed on a rough horizontal table. The table is gradually tilted until the particle begins to slip. The block is on the point of slipping when the table is inclined at an angle of 41º to the horizontal. Find the coefficient of friction.

Page 10: Forces

3 minutes later the answer appears

Resolving forces parallel to the planeF = 3 g sin 41º = 19.3 N

Resolving forces perpendicular to the planeR = 3 g cos 41º = 22.2 N

But the particle is on the point of sliding so

F≤R

19.3 = 22.2

= 0.869 (to 3 s.f.)

N.B. You could have used = tan-1 41º

Page 11: Forces

More questions

Click to advance slide

1. A block of mass 6 kg will move at constant velocity when pushed along a table by a horizontal force of 24 N. Find the coefficient of friction between the block and the table.

2. A puck of mass 0.1 kg is sliding in a straight line on an ice rink. The coefficient of sliding friction between the puck and the ice is 0.02. Find the resistive force due to friction and then find the speed of the puck after 20 seconds if its initial speed is 10 ms-1.

Click here for answers

Page 12: Forces

3 A particle of mass 1 kg is projected at 5 ms-1 along a rough horizontal surface. The coefficient of sliding friction is 0.3. Assuming that F = N, how far does the particle move before coming to rest?

4 A gymnast of mass 80 kg is sliding down a vertical climbing rope with constant speed. The coefficient of friction between his hands and the rope is 0.3. Calculate the total normal contact between his hands and the rope. State any assumptions you make.

5E A particle rests on a rough plane inclined at an angle f to the horizontal. When an additional force P is applied as shown, the particle slides down the slope at constant speed. Show that P = mg(sin - cos ) where m is the mass of the particle and m is the coefficient of friction. If the additional force is trebled to 3P, and the particle now moves up the plane with constant speed, show that tan = 2.

Click here for answers Click to advance slide

Page 13: Forces