{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# T1a4s - Assignment 4 solutions(Fall term Physics 102 4.37(a...

This preview shows pages 1–2. Sign up to view the full content.

4.37 (a) Since the crate has constant velocity, 0 x y a a = = . Applying Newton’s second law: , or cos20.0 0 x k x Assignment 4 solutions (Fall term Physics 102) F F f ma Σ = ° − = = ( ) 300 N cos20.0 282 N k f = ° = and , or sin 20.0 0 y F n F w Σ = ° − = ( ) 3 300 N sin 20.0 1000 N 1.10 10 N n = ° + = × . The coefficient of friction is then 3 282 N 1.10 10 N k k f n µ = = = × 0.256 . (b) In this case, , so . The friction force now becomes sin 20.0 0 y F n F w Σ = + ° − = sin 20.0 897 N n w F = ° = ( )( ) 0.256 897 N 230 N k k f n µ = = = . Therefore, cos20.0 x k x w x F F f ma g Σ = ° − = = a and the acceleration is ( ) ( ) [ ] ( ) 2 300 N cos20.0 230 N 9.80 m s cos20.0 1000 N k F f g a w ° − ° − = = = 2 0.509 m s 4.41 The normal force acting on the crate is given by cos n F mg θ = + . The net force tending to move the crate down the incline is sin s mg f θ , where s f is the force of static friction between the crate and the incline. If the crate is in equilibrium, then sin 0 s mg f θ = , so that sin s f mg θ = . But, we also know ( ) cos s s s f n F mg µ µ θ = + .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

T1a4s - Assignment 4 solutions(Fall term Physics 102 4.37(a...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online