The coefficient of static friction between a 2.70 kg crate and a 35.0° incline is 0.320. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

(If the image is unclear, this is what a diagram would look like: a 2.7 kg crate on a 35 degree incline, and the applied force must be perpendicular to the crate on the incline.)

## Answers (2)

k = coefficient of static friction =0.32

(mgcos35° + F)*k = mgsin35

F =( mgsin35) /k - mgcos35

F = mg(1/k*sin35 - cos35)

F = 2.7*9.8( 3.125*0.5736 - 0.8191) = 25.76N

2.70*9.81*sin 35 = 0.320*[(2.70*9.81*cos 35)+F], where F is the required force applied normal to the incline on the mass. Or

15.19 = 0.320*2.70*9.81*cos 35 + 0.320*F or

15.19 = 8.476 + 0.320*F or F = (15.19 - 8.476)/0.320 = 20.9889 or 21 N