I am pushing down on my lawnmower an angle θ=25 degrees below the horizontal with a force 750 N where the static friction coefficient between the at surface and the trunk is μs=0.75. What is biggest lawnmower I will be able to move?
Here's my attempt:
The sum of all components in the y direction is N-mg-750sin(25)=0, which means N=mg+750sin(25).
The sum of all components in the x direction is 750cos25-fs=0, with fs being the static fiction. We know that fs=μs*N
Combining what we know is N and our previous equations together we have the sum of all x components being
750cos(25)-0.75(mg+750sin(25))=0 -->679.73-7.35*m-237.72=0 -->7.35m=442.01 -->m=60.14 kg
Am I understanding the use of static friction here correctly? I'd appreciate any input I can get.
Let be $F=750\,\mathrm N$, $\mu_s=0.75$, $g=9.8\,\mathrm{m\,s}^{-2}$ and $\theta=25^\circ$ $$\left\{\begin{align} F\cos\theta-f_s&=0\\ F\sin\theta+mg-F_n& = 0\\ f_s&=\mu_sF_n \end{align}\right. $$ We have $f_s=\mu_s(F\sin\theta+mg)$ and $F\cos\theta=\mu_s(F\sin\theta+mg)$ that is $$ m=\frac{F}{g}\left(\frac{1}{\mu_s}\cos\theta-\sin\theta\right)=60.14\,\mathrm{kg} $$