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A steel washer suspended from the top of a crate sliding down a ramp.

  • #1
Question:
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 36^\circ above the horizontal. The crate has mass 156 {\rm kg}. You are sitting inside the crate (with a flashlight); your mass is 64 {\rm 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 66^\circ with the top of the crate.

So far, I've just figured out:

y: n=mcrate+persongcos(theta)

x: fk-mcrate+persongsin(theta)=ma
u(mcrate+persongcos(theta))-mcrate+persongsin(theta)=ma
u=[a+gsin(theta)]/(gcos(theta))

so using that I got g=9.8m/s^2, theta=36, mcrate+person=220kg, but I don't know what a is...

my question is... Do we have to use the additional information (the washer) to find a? if so, how would we solve for a?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
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welcome to pf!

hi gigazard332! welcome to pf! :smile:

(have a degree: ° and a theta: θ and a phi: φ and try using the X2 tag just above the Reply box :wink:)

this question has deliberately given you a lot of information so as to test whether you can see what is relevant and what isn't …

so i won't spoil it by telling you! :biggrin:

notice that you haven't been given the coefficient of friction, and instead you've been given the angle of φ = 66° for the string …

assume that the acceleration is a, calculate φ, then put φ = 66° to get an equation for a …

what do you get? :smile:
 

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