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.

  • 2
    $\begingroup$ 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. $\endgroup$ Jun 2, 2014 at 19:14
  • $\begingroup$ Do you mean $f_k=\mu*F_n*cos \theta$ ? $\endgroup$
    – CharlieK
    Jun 2, 2014 at 19:26
  • 3
    $\begingroup$ More like $f_k = \mu \times (mg +$ the contribution to the normal force from pushing). $\endgroup$ Jun 2, 2014 at 20:02
  • $\begingroup$ $\sum F=0$ isn't Newton's first law. $\endgroup$
    – Ken
    Nov 11, 2020 at 23:31

3 Answers 3


Horizontal component of force

$$F_x=F\cos32^\circ-f$$ Vertical component of force

It must be zero as block will remain intact with ground. $$F_y=N-mg-F\sin32^\circ = 0$$ Normal force (due to ground)

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

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

The horizontal component must be infinitesimally small positive force. In other words, you can equate it to zero to get $F$. $$F_x=F\cos32^\circ-\mu (mg + F\sin32^\circ)\to0^+$$ $$\large F=\frac{\mu mg}{\cos32^\circ-\mu \sin32^\circ}\approx100\,\text{N}\left(98.88\right)$$ Free body diagram

  • $\begingroup$ @Davidmh is it better $\endgroup$
    – RE60K
    Jun 9, 2014 at 3:59
  • $\begingroup$ it is much more clear now. $\endgroup$
    – Davidmh
    Jun 9, 2014 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

  • $\begingroup$ I did it using $f_k=μ∗F_n∗cosθ$ and still my answer was wrong. $\endgroup$
    – CharlieK
    Jun 2, 2014 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, 2014 at 19:59
  • 1
    $\begingroup$ 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, 2014 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, 2014 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 $F\cos(32) −f_k = 0$. The x-component of $F$ is $F_x = F\cos(32)$. Writing $F_x \cos(32)$ doesn't make logical sense.

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


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