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The component of the force of gravity, pointing parallel down the incline, is:

Fd = m g sin(θ)

The maximum force of friction is:

Fd = µ m g cos(θ)

The net force is:

Fnet = m g sin(θ) - µ m g cos(θ)

Fnet = m g (sin(θ) - µ cos(θ))

The acceleration is:

Fnet = m a

a = Fnet / m

a = g (sin(θ) - µ cos(θ))

a = 9.8 (sin(30) - 0.15 cos(30))

a = 3.6 m/s²

Note that, in order to start moving, Fd must be bigger than Ff. If it is smaller, then friction will 'adust itself' to be equal to Fd, thus making the net force zero, thus keeping the skier stationary.

What help do you need? Are you not able to click the equation required in this question.REMEMBER one thing EQUATIONS OR CONSTRAINTS are required to crack most of questions in MECHANICS [specially in questions where acceleration or friction or any other force calculation is required].

The only key to mechanics is to understand,feel and visualize the problem clearly,then make equations or constraints .remember number of equations depend upon the number of variable(what is required to be found out)

This is the most primary level of question in FRICTION engaging slope or inclination.NO WORRIES!!TAKE IT EASY.

The first line reads ''a skier has just begun descending a 30 degree slope'',means it is the inclined FREE FALL OF SKIER because skier has just begun descending means component of gravitational force along the surface of slope pulls him ahead along surface of slope. ------ a

second line speaks of kinetic friction ,are you obstructed because of this?You saw the word friction and you are destroyed,then kinetic added to it ,makes you worse ,is it???

Do you really know ,first of all ,as to what is friction,do you have it's sole meaning cleared ?because without it you would not be able to move ahead in this question.

SEE

FRICTION is force acting between the 2 bodies in contact acting in direction opposite to relative motion of 2 bodies.

HERE the direction of relative motion is the direction of descending of skier because the 2nd body i.e.slope is at rest,it is fixed and the only moving body is skier ,so net relative motion is motion of skier.

NOW according to the definition of friction the direction of friction force is opposite to the relative motion,therefore the direction of kinetic friction is against the direction of propagation of skier.------- b

Tell me were you thinking why is it kinetic? and got a problem due to this ,because when any motion or relative motion occurs friction force generated is termed as KINETIC and when no motion occurs is termed as STATIC.

FINE

now we will make an equation so as to solve this question=>

according to 'a' and ' b',applying the NEWTON'S SECOND LAW OF MOTION=>

NET FORCE ACTING OVER SKIER = ITS MASS X ITS ACCELERATION [newtons second law of motion states that net force f acting on a body = m x a} ------ c

now let us summarize all the forces acting over skier in its direction of motion;

firstly ,it is the component of gravity acting along the surface of slope,assume mass of skier as 'm',then this force is nothing but mgsin 30 degree.

Now how mgsin 30 degree ??

see, if you draw the FREE BODY DIAGRAM of skier,then vertical force is mg which makes an angle

60 degree with slope,isn't it?there fore if you assume a right angled triangle with mg vector as hypotenuse ,component of mg along slope as base and component of mg NORMAL to the slope as perpendicular then you will find that component of mg along the slope will be mgsin30 degree and component of mg NORMAL to slope will be mg cos 30(USE OF BASIC TRIGONOMETRY)

[WHICH IS VERY CLEAR FROM THE FIGURE ILLUSTRATED ,FROM THE GIVEN BELOW LINK]

NOW we know that for coefficient of friction as Uk ,friction force is Uk x N,WHERE N is the normal force acting on body.here N = mgcos 30 degree as calculated above.thus frictional force acting against the skier = .15 x mg cos 30 degree.Hence the friction force acting in direction of motion of skier is - .15 mg cos 30 degree.

now we will make the equation so as to solve this question=>

according to 'a' and ' b',applying the NEWTON'S SECOND LAW OF MOTION=>

NET FORCE ACTING OVER SKIER = ITS MASS X ITS ACCELERATION [newtons second law of motion states that net force f acting on a body = m x a} ------ c

c can be rewritten as

mgsin 30 degree - .15mgcos 30= m x a

g(1/2 - .15xsquare root of 3/2)=a

10(1/2 - .15 x .8)=a

10 x .38= a

a=3.8 m/s^2

basically you needed to crack the equation that was the question actually asking

STEP WISE SUMMARY;

STEP NO. 1

analyse the situation i.e. skier is freely falling under the effect of gravity and frictional force.

STEP NO . 2

MAKE THE FREE BODY DIAGRAM AS SHOWN IN FIGURE IN THE LINK BELOW.

STEP NO. 3

write the equation as in accordance with the FREE BODY DIAGRAM.

HOPE YOU ARE CLEAR WITH YOUR PROBLEM