Problem with friction force


by Hernaner28
Tags: force, friction
Hernaner28
Hernaner28 is offline
#1
Mar25-12, 01:34 PM
P: 263
1. The problem statement, all variables and given/known data

The surfer (70kg) is standing over a board which has an angle of 10 with the horizontal. A wave acelerates him to the right at 1.0m/s^2. There is a force that prevents the surfer from sliding over the board.


2. Relevant equations
Calculate that force. What's the origin of this force?


3. The attempt at a solution

What I did was to analize the surfer only and since there's horizontal aceleration to the right then there's a wave force in horizontal direction to the right. So I just analized it in axis x given the faact that at axis y it doesn't move. So the equations:





Then I calculated the weight at axis x:



So the friction force has to be the difference among these two and in the direction of the wave force and it would be 50N but the option is not in the moddle multiplechoice I'm doing.

Thanks!!
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francesco85
francesco85 is offline
#2
Mar25-12, 04:02 PM
P: 62
Quote Quote by Hernaner28 View Post
1. The problem statement, all variables and given/known data

The surfer (70kg) is standing over a board which has an angle of 10 with the horizontal. A wave acelerates him to the right at 1.0m/s^2. There is a force that prevents the surfer from sliding over the board.


2. Relevant equations
Calculate that force. What's the origin of this force?


3. The attempt at a solution

What I did was to analize the surfer only and since there's horizontal aceleration to the right then there's a wave force in horizontal direction to the right. So I just analized it in axis x given the faact that at axis y it doesn't move. So the equations:





Then I calculated the weight at axis x:



So the friction force has to be the difference among these two and in the direction of the wave force and it would be 50N but the option is not in the moddle multiplechoice I'm doing.

Thanks!!

Hi, in my opinion the answer is the following: in the (accelerated) frame where the man is at rest the following forces act on the man: gravity (m g), inertial force (-m a), the reaction of the table which is perpendicular to the plane of the board (R) and and friction (F) which is parallel to the board; if you sum (vectorially of course) all these force you should obtain zero in order to say that the man is in equilibrium over the board; you have then two equations in two unknowns (F and R); solving the system I obtain

[itex]F=m(g [/itex] sin[itex](\theta) +a[/itex] cos[itex](\theta))[/itex]

substituting [itex]\theta=10[/itex], [itex]a=1\frac{m}{s^2}[/itex], [itex]m=70Kg[/itex], I obtain
F≈188N
Hernaner28
Hernaner28 is offline
#3
Mar26-12, 08:06 AM
P: 263
Oh yes! You're right! I forgot that the inercial force is negative! Thanks!


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