- #1
cak
- 15
- 0
Here is the problem I'm having trouble with...
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?
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?