Is a 5° Ramp Slope Too Steep for Pushing a 30kg Cart?

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SUMMARY

A slope of 5° is too steep for pushing a 30kg cart, as the net force required exceeds the acceptable limit of 50N. The calculations show that the frictional force (Ff) is 29.88N, and when combined with the force required to push the cart (Fp), the total force (FNet) reaches 55.88N. This value surpasses the threshold, confirming that the ramp's incline is inappropriate for customer use. The correct gravitational force should be calculated using g=9.8m/s², leading to a more accurate normal force value.

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Homework Statement


In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to push grocery carts up the ramps and it can't be too difficult. An engineer has done a survey and found that no one complains if the force required is no more than 50N. Will a slope of θ=5° be too steep, assuming the cart has a mass of 30kg? Assume μk=0.10

Homework Equations


Ff = μkFn

The Attempt at a Solution



Yes, it is too steep. Why? I have no idea...

Below is my diagram...forgive the fact that Fp is not parallel to the ramp's surface. It's supposed to be, but it's the best I can do in MS Paint right now.

Ff = μk * FN
= 0.1 * 299N
= 29.88 N

FNet = Ff + Fp = 55.88N (greater than the required 50N limit)

Is that right? I've been sitting here looking at this stupid thing for over an hour and I think I may have just gotten it. The one thing I don't like about these textbooks is that they give the answers to selected problems but no explanation. Just because my answer here is over 50N doesn't necessarily mean that it's right. So is it?
 

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TrpnBils said:
FNet = Ff + Fp = 55.88N (greater than the required 50N limit)

Is that right? I've been sitting here looking at this stupid thing for over an hour and I think I may have just gotten it. The one thing I don't like about these textbooks is that they give the answers to selected problems but no explanation. Just because my answer here is over 50N doesn't necessarily mean that it's right. So is it?

It is over 50 N, the numerical result is almost right, and yes, the slope is too steep. You did it well, why do you not trust in yourself? :smile: one small mistake: mg = 30*9.8cos5=293 instead of 299. What value have you used for g? ehild
 
Last edited:
According to the diagram he used g=10m/s/s. May have been told to use that. The 299 is mg.sin(5deg). The reasoning shown is sound - needs to draw the friction in the free-body diagram but that's a quibble. Just a tidy up.

One of the ways to gain confidence in your calculations is to keep track of your reasoning at each step. If your initial attempt looks kinda all over the place it can help to pretend you are explaining how to do it to someone else (like us :) ) and you need to keep the explanation simple and short as you can.
 

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