How Much Force to Lower Two Boxes Down a Ramp at Constant Speed?

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Homework Help Overview

The problem involves lowering two boxes down a ramp at a constant speed, requiring the calculation of the force exerted and the friction force on the upper box. The scenario includes specific coefficients of friction and a defined speed for the boxes.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss summing the weights of the boxes to find the necessary force and question how to determine acceleration given the constant speed. There are attempts to calculate forces and friction, with some participants seeking verification of their results.

Discussion Status

Some participants have provided calculations for the forces involved, while others have pointed out potential errors in reasoning and calculations. There is an ongoing exploration of the problem with no explicit consensus on the correctness of the results yet.

Contextual Notes

Participants are encouraged to show full calculations for verification, and there are reminders about proper notation and checking assumptions in their work.

truthlies217
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Homework Statement


can someone help me solve this problem .. You are lowering two boxes, one on top of the other, down the ramp shown in Figure 5.53 by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800. (a) What force do you need to exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box?

Figure here > http://www.flickr.com/photos/53947633@N06/4988606991/

Homework Equations



Sum of F = m a

The Attempt at a Solution



i just sum the weight of the two boxes to find the force for part one and i still need the acceleration so how can i find it ?
 
Last edited:
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welcome to pf!

hi truthlies217! welcome to pf! :smile:
truthlies217 said:
i just sum the weight of the two boxes to find the force for part one …

yup! :biggrin:
and i still need the acceleration so how can i find it ?

hmm, let's see …

"Both boxes move together at a constant speed of 15.0 cm/s." …

so the acceleration is … ? :wink:
 
opss .. ><

Ok then i found F= 346.05 N for part a and fk= 627.2 for part B the direction is 207.2 degree .. I'm not sure is it right ?
 
Last edited:
hi truthlies217! :smile:

if you want us to check your results, you need to show us your full calculations :wink:
 
part A :

mg= 32+48= 80 kg*9.8 = 784N
the angel = shift tan ( 2.50/4.75) = 27.7 degree
/x =784+Cos27.7 + F - Fk = 0.0
/y =784+Sin27.7+n=0.0
so n=784 N
F=784+Cos27.7-(784*0.444) = 346.05 N

part B :

fk=un = (0.800*784)=627.2 N
direction = 27.7 + 180 = 207.7 degree ..
 
truthlies217 said:
/x =784+Cos27.7 + F - Fk = 0.0
/y =784+Sin27.7+n=0.0
so n=784 N

(please don't use + to mean "times" :frown:)

no, your cos and sin are the wrong way round (you should always check by asking whether it gives the right result if the angle is zero)

and n = 784 N doesn't follow from the previous line anyway
 
i just found my mistakes .. thanks for the help :)
 

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