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I ran into this problem studying exercises for my exam. The coefficient of kinetic friction is 0.5, the mass is 7.3kg and the force is applied at an angle of 45°. The problem says to find the force F that will move the box at a constant speed along the horizontal floor. What I don't understand is the 'constant speed', this means not accelerating but the body has to at least accelerate a little to even start moving. The question also doesn't specify the value of this constant speed. Am I right to think it is impossible to solve this without more information? enter image description here

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    $\begingroup$ The question assumes that the body is already moving with constant speed and so the net horizontal force on the body is zero. Note that because the force$F$ has a vertical component the normal reaction on the block is not $mg$ $\endgroup$
    – Farcher
    Oct 5, 2016 at 8:41
  • $\begingroup$ @Farcher if I understand correctly, Fsin45+R=73N, where R is normal reaction and taking g=10m/s^2. And Fcos45=f if the net horizontal force is zero. Am I going in the right direction? $\endgroup$
    – lekarane
    Oct 5, 2016 at 8:59

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Why complicate things when you DO NOT need the value of the speed to solve it ?

Constant speed here is assuming that the box is already in a state of a motion of a=0 and v!=0 at t=0. From t=0 onwards, F and the frictional force, f=µ·mg. (here 36.5) and F (at an angle θ=45°) start acting on the block.

Direction of F·cosθ is opposite to that of f, which can be used to cancel the effect of friction.

To maintain the constant motion, F·cosθ should be equal to f ie. F/sqrt(2) = 36.5

F= 36.5 · 1.414 = 51.611

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  • $\begingroup$ v!=0 at t=0? What does that mean? $\endgroup$ Oct 5, 2016 at 14:05
  • $\begingroup$ Welcome on Physics SE :) Thank you for your contribution. You might want to visit the help centre to see how to better typeset formulas. $\endgroup$
    – Sanya
    Oct 5, 2016 at 15:45
  • $\begingroup$ See here for help on math formatting. In short, enclose math with dollar signs. For example $ x^2+1$ renders as $x^2+1$. Next use the special keywords for fractions (example $\frac{x}{2}$ is \frac{x}{2}), square root (example $\sqrt{x}$ is $\sqrt{x}$ and finally for functions use roman (non italisized characters). For example $\sin^2(x) +\mathrm{g}(x)$ is $\sin^2(x) +\mathrm{g}(x)$. $\endgroup$ Dec 22, 2017 at 15:30

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