# Friction force acting on a block

I have attached a picture file and a free-body diagram file for this problem.

Determine the friction force acting on the block shown when the angle is 30 degrees and P=200N.

us=0.3 (static friction)

uk=0.2 (kinetic friction)

The free-body diagram is how I started this solution. My equations became:

Fx=0=200-Ncos(60)-fcos(30)

Fy=0=-1000-fsin(30)+Nsin(60)

When I tried to substitute in uN for f I got two different answers for N. I tried this method for usN and ukN. Where am I going wrong?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

If you assume in advance that the block will not move, then the equations you wrote down are correct. But they are not written down in a convenient coordinate system. Let's work with your equations first.

The assumption that the block will not move can be incorrect. However, this is not a problem, because you can use your equations to solve for both the normal force N and the friction force f. The value for f you get must then be interpreted as the friction force that is necessary for the block to remain at rest. You then calculate the ratio of f and N to see if it exceeds us. If not, then the block will indeed remain at rest. If the ratio exceeds us, then the block will accelerate. The friction force of f you calculated then cannot be supplied. Instead the friction force will be uk times the normal force N.

You wrote down the equations:

Fx=0=200-Ncos(60)-fcos(30) (1)

Fy=0=-1000-fsin(30)+Nsin(60) (2)

Here you take f to be positive if it points downward the slope. The forces are given in units of Newton, but we'll omit the "N" to avoid confusion with the normal force denoted by N. It is convenient to express the equations in terms of one angle only:

Nsin(30) + fcos(30) = 200 (3)

-fsin(30)+Ncos(60) = 1000 (4)

Multiply (3) by cos(30) and (4) by sin(30) and subtract the two equations. Then the normal force N drops out. Using Sin^2(30) + Cos^2(30) = 1 you obtain:

f = 200 cos(30) - 1000 sin(30) = -326.8 (5)

If you multiply (3) by sin(30) and (4) by cos(30) and add the two equation you obtain:

N = 200 sin(30) + 1000 cos(30) =966 (6)

The fact that f is negative means that it is directed along the slope in the upward direction.

The absolute value of f/N is 0.338 which is larger than larger than us, so we can conclude that the block will not remain at rest. The block will accelerate which means that the net force in both the x and y directions will not be zero. This then invalidates the equations (1) and (2). However, the normal force of 966 that we just obtained is still correct, as I'll explain below. This means that the friction force will be uk times 966 which is 193, and this is directed upward the slope.

To see that the computed value for N of 966 is correct, let's use a different coordinate system. Let's define the x-direction as the direction along the plane in the upward direction and the y direction is the direction normal to the plane. We know that the block will accelerate, so the net force on the block is not zero. But since the block has to move along the slope, the component of the acceleration in the y-direction must be zero. This means that the net force in the y-direction must also be zero. This yields the equation:

-200 sin(30) - 1000 cos(30) + N = 0 (7)

This is the same as Eq. (6).

In these type of problems it is convenient to choose your coordinates like this right from the start!

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