A factory worker pushes a 29.7kg crate a distance of 5.0m along a level floor at constant velocity by pushing downward at an angle of 32$^\circ$ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

What magnitude of force must the worker apply to move the crate at constant velocity?

So because constant velocity \begin{align}\sum F&=0\\ F_x\times \cos{(360^\circ-32^\circ)} - f_k &= 0\\ F_x\times \cos{(360^\circ-32^\circ)} - (mg)\times 0.24 &= 0\\ F_x\times \cos{(328^\circ)} - 69.8544\,\mathrm N &= 0\\ F_x &= 69.8544\,\mathrm N/\cos{(328^\circ)}\\ F_x&=82.3708\,\mathrm N \end{align}

That turns out to be wrong, the expected answer is $f_k$ the force of kinetic friction. Why? Shouldn't I be able to use $\sum F=0$ in this problem to find the answer?

I also tried to find $\sqrt{F_x^2+F_y^2} -f_k$ but that didn't help either.

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    $\begingroup$ You are using only $mg$ as normal, but the force that is acting at an angle would also add up a part in the normal reaction! Correct your normal force and you must get desired answer. $\endgroup$ – Rijul Gupta Jun 2 '14 at 19:14
  • $\begingroup$ Do you mean $f_k=\mu*F_n*cos \theta$ ? $\endgroup$ – CharlieK Jun 2 '14 at 19:26
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    $\begingroup$ More like $f_k = \mu \times (mg +$ the contribution to the normal force from pushing). $\endgroup$ – NeutronStar Jun 2 '14 at 20:02

Horizontal component of force


Vertical component of force

It must be zero as block will remain intact with ground. $$F_y=N-mg-F\sin32^o = 0$$

Normal force (due to ground)

What you are missing is that you haven't added the increase in $N$ due to $F\sin32^o$ $$N = mg + F\sin32^o$$

Frictional force

$$f=\mu N=\mu (mg + F\sin32^o)$$

The horizontal component must be infinetstimally small positive force.In other words you can equate it to zero to get $F$. $$F_x=F\cos32^o-\mu (mg + F\sin32^o)\to0^+$$ $$\large F=\frac{\mu mg}{\cos32^o-\mu \sin32^o}\approx100N\left(98.88\right)$$ Free body diagram

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  • $\begingroup$ It would be better if you add some explanations. $\endgroup$ – Davidmh Jun 7 '14 at 13:36
  • $\begingroup$ @Davidmh is it better $\endgroup$ – RE60K Jun 9 '14 at 3:59
  • $\begingroup$ it is much more clear now. $\endgroup$ – Davidmh Jun 9 '14 at 7:22

The normal force is not only provided by the weight of the body but also by the force you have applied. Take it into consideration and you will get the answer

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  • $\begingroup$ I did it using $f_k=μ∗F_n∗cosθ$ and still my answer was wrong. $\endgroup$ – CharlieK Jun 2 '14 at 19:56
  • $\begingroup$ Other option is to add $F_y$ to the object's weight, thus increasing the normal force. Is this second method right? $\endgroup$ – CharlieK Jun 2 '14 at 19:59
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    $\begingroup$ the net normal force would be equal to the y component of the applied force plus the weight of the object $\endgroup$ – Normie Jun 2 '14 at 20:02
  • $\begingroup$ I did this: $(Fx*tan(328)+mg)*.24$ and still got it wrong. I don't believe what you are saying. Unless you can prove to me why it is $F*sin(328)$ and not tan. I use tan because of $tan\theta=y/x$ $\endgroup$ – CharlieK Jun 3 '14 at 7:42

I'm going to just use $32^∘$ below; it doesn't make a difference. Your equation isn't correct. You should have $F_x−f_k = 0$ or $Fcos32∘−f_k = 0$. The x-component of F is $F_x = F.cos32^∘$. Writing $F_x.cos32^∘$ doesn't make logical sense.

Why? Shouldn't I be able to use $∑F=0$ in this problem to find the answer? Yes, you can.

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