Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Friction Problem of a Crate Sliding Down a Ramp
Reply to thread
Message
[QUOTE="BBA Biochemistry, post: 5796123"] [h2]Homework Statement [/h2] A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of 38 ∘ above the horizontal. The crate has mass 239 kg . You are sitting inside the crate (with a flashlight); your mass is 59 kg . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of 71 ∘ with the top of the crate. What is the coefficient of kinetic friction between the ramp and the crate? Note: u=coefficient of friction [h2]Homework Equations[/h2] Newton's Second Law: F=ma Kinetic Friction: Fk=uN [h2]The Attempt at a Solution[/h2] I have defined my xy-coordinate system with the x-axis along the ramp (negative going up the ramp) and the y-axis perpendicular to the ramp. Forces on the crate: Fx= Mgsin38 - uMgcos38=ma. (Eq. 1) There is no net force in the y-direction on the crate. Gravity is acting in the positive x-direction (for how I defined the xy-coordinates), and friction acts opposite to the direction of motion, so it is negative. So Fx= (x-component of gravity)-(friction force). Friction force is equal to uN, and the Normal force is equal to the y-component of gravity (which is Mgcos38). The washer is the trickier part to me. It seems that I need to use the washer to find the acceleration. Here are the equations I have for the forces on the washer. Note: T=tension of rope on washer. Also, little m denotes mass of washer, whereas big M denotes mass of crate. Forces on the washer: Fx= Tcos71+mgsin38. (Eq. 2) Both gravity and the tension act on the string in the positive x-direction (I think). Since m and T are not given, they somehow need to be eliminated, so I defined a second equation. Forces in the y-direction on washer: Fy= Tsin71 - mgcos38=0. (Eq. 3) Since the washer is not moving in the y-direction, the sum of the forces is 0. I then solved Eq. 3 for T, getting: T= [mgcos(38)] / [sin(71)] I plugged this into Eq. 2, getting: Fx= [mcos(38)cos(71)] / [sin(71)] - mgsin(38)=ma. Since every term has m, they can all cancel out, leaving: Fx= [cos(38)cos(71)] / [sin(71)] + gsin(38) = a = 6.3 m/s^2 Now that acceleration is known, there is only one unknown in Eq. 1, the friction coefficient. Mgsin38-uMgcos38= Ma (This is just Eq. 1 restated) Solving for u: Since M is in every term, I can cancel out M: gsin38-ugcos38=a. Rearranging the equation to isolate u: gsin38-a=ugcos38 Divide both sides by gcos38: tan38- [a] / [cos38] =u. u= -0.35. I figure the negative is since it's just in the opposite direction. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Friction Problem of a Crate Sliding Down a Ramp
Back
Top