Force Problem

[EvoN1020v]

Thanks 5614. I got another challenging physcis question for you guys:

 

A 40 kg boy works at his dad's hardware store. One of the boy's jobs is to unload the delivery truck. He places each package on a 30 degrees ramp and shove it up the ramp into the storeroom. He needs to shove the package with an acceleration of at least 0.1 m/s^2 in order for the package to make it to the top of the ramp. One day the ground is wet with rain and he's wearing slick leather-soled shoes. The coefficient of static friction between his shoes and the ground is 0.25. The largest package of the day is 15 kg, and it's coefficient of static friction on the ramp is 0.40. Can he give the package a big enough shove to reach the top of the ramp without his feet slipping?

[5614]

Well lets see:

 

He needs to push at a=0.1

 

Biggest mass = 15

 

therefore F = ma = 15 * 0.1 = 1.5N

 

Now can I assume that he pushes up at 30 degrees or does his line of force act horizontally? That makes a difference. If he pushes it at 30 deg then he needs F = 1.5 but if he pushes horizontally he would need a force which gives a component of the force up the slope of 1.5N

 

See the difference? Does he need to supply the 1.5N or supply something which has a component of 1.5N ?

 

Anyway once you know the force he needs to push then remember that if he pushes the load with a force F then the load pushes him with a force F (in the opposite direction). Include friction, work out the overall force required to move the box. Use [math]F = \mu R [/math] to see if the force he has to supply is bigger than the frictional force which is available. If friction is bigger than he can move it, if friction is smaller than the force he is applying then he will slide.

[EvoN1020v]

One thing: I mistype the value of acceleration, it should be [math]1.0 m/s^2[/math] instead. Therefore, the applied force ([math]F_{app}[/math]) would be 15 N.

 

First thing I did: Find the friction Force when the boy push the 15 kg box.

 

[math]\sum F_x = F_{app} - F_f - (F_g)_x = ma[/math]

[math]F_f = F_{app} - (F_g)_x - ma[/math]

 

[math]F_{app}[/math] and [math]-ma[/math] cancel each other, so that remains [math](F_g)_x[/math] which is [math]mgsin30[/math].

 

The friction force is: [math]73.575 N[/math].

 

What I do next? I know that the friction force is greater than the applied force but, he have a mass of 40 kg, so I'm not sure how to join that reference to figure out my answer?

 

Maybe I should figure out the [math]F_{fsmax}[/math] for his shoe to the ground? Any help here please?

[5614]

I haven't properly read your post, just the question, will reply properly tomorrow but if it helps then to involve the boy's mass... well I dunno what you did without it! (can't stop to read it now) but the boys mass will effect the maximum amount of friction he can have.

 

Frictional forces are [math]F= \mu R[/math] where R is his reaction, which will be equal to his weight. Weight = mg. That's where 'm' (mass) comes into it.

 

Will look at the prob more tomorrow.

[EvoN1020v]

never have seen [math]F = \mu R[/math] before?

 

I know that the boy's weight is [math]392.4 N[/math].

[5614]

Doubley posty, oops!

[5614]

You have seen [math]F = \mu R[/math] before! How do you work out the force which can be supplied due to friction? I think you do know it but maybe just haven't ever had it written out like that.

 

The force ([math]F[/math]) is equal to the coefficient of friction ([math]\mu[/math]) multiplied by the reaction ([math]R[/math]).

 

If the boy's mass is 40kg, then his weight is 392N (I use g as 9.8) and the coefficient of friction is 0.25 then the maximum value of friction (which is a force) is [math]F_{friction} = \mu_{coefficient-of-friction} R_{reaction} = 0.25_{given-in-the-question} * 392_{mass * gravity} = 98N[/math]

 

Now remember Newtons 3rd Law, if the boy pushes the box with a force F then the box pushes the boy with a force F in the opposite direction.

 

So now quite simply if he has to push with more than 98N then he will be being pushed with more than 98N and so he will slide, because the greatest force friction can supply is 98N. If he's pushing the box with >98N then he's being pushed with >98N which will overcome friction and so he will slide.

 

The problem to me is what force does he apply? Do you follow what I was saying at the end of post #27? Is he applying a force horizontally or is he applying the force up the slope? That makes a difference to the answer.

 

You know how to work out the frictional force on the box, it's the same method for the boy.

 

The maximum force due to friction on the box I get as [math]F_{friction} = \mu R = 0.25 * 73.575_{mgsin \theta = 15 * 9.8 * sin30 = 73.5 (rounded)} = 18.394N[/math] so now you need to account for the force required to push the box (which I'm not certain on, see above) and then add the friction.

 

Is it getting clearer?