- #1

cak

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The coefficient of static friction between the 32 kg crate and the 35.0° incline of Figure 4-34 is 0.2. What is the minimum force, F, that must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

I did a force diagram and figured out the variables:

1. I got normal Force as 257.152

2. force of friction Fk=µk*N = .2 * 257.152 = 51.43

3. perpendicular force = 32 * 9.8 * cos(35) = 256.89

4. parallel force = 32 * 9.8 * sin(35) = 179.87 = 179.87

5. Net force=parallel force - force friction = 179.87 - 51 = 128.49

But that isn't right. Can someone help me?