Static Friction - starting motion

AI Thread Summary
To determine the force needed to start moving a 32kg crate on a rough floor with a static friction coefficient of 0.50, the equations for forces in both the x and y directions must be established. The normal force is not simply the weight of the crate; it is affected by the vertical component of the applied force due to the angle of 21 degrees. The correct approach involves calculating the net forces and ensuring that the static friction force is equal to the applied force in the x-direction. The initial calculations provided were incorrect because they did not account for the angle's effect on the normal force. Properly setting up the equations will lead to the correct solution.
surfinusa555
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To move a large crate across a rough floor, you push on it with a force at an angle of 21 degrees below the horizontal, as shown in the figure:
http://img227.imageshack.us/img227/4950/walker67ac0.jpg

Find the force necessary to start the crate moving, given that the mass of the crate is 32kg and the coefficient of static friction between the crate and the floor is 0.50.

<br /> F_x = \mu_s * F_N<br />

Here is what I did, and got the wrong answer:
<br /> M_c = 32kg \\*<br />

<br /> \mu_s = 0.50<br />

<br /> F_x = \mu_s * F_N<br />

So...
<br /> F_x = \mu_s * (M_c) * (a_g)<br />

<br /> F_x = 0.5*(32kg)*(9.8 m/s^2)<br />

<br /> F_x = 156.8N<br />

So...
<br /> Cos(21) = (156.8 N)/C<br />

C = 167.96 N

That answer is wrong... so what am I doing wrong?
 
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Re-examine your acceleration in the X direction.
 
I'm sorry, I still don't understand...
are you saying that the normal force is not equal to (32)*(9.8)?
 
First write the equations for \Sigma{F_x} = ma_x and \Sigma{F_y} = ma_y.

This is the most important step. If you have the equations right, then the problem is easily solved.
 
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