A man pulling sled of his daughter by a massless rope, climbing a snowy hill whose slope is equal to 15 °. Considering that the mass of the sled is $4Kg$, the girl's $26Kg$ and $\mu _c = 0,25$, calculate the work done by the tension of the string after walking with constant speed a distance of $130m \Delta x$
So, I know:
$\varphi=15°; \alpha =30°; m_d=26Kg; m_s=4Kg; \mu _c =0,25; d=130m$
So, the equations will be:
Weight in Y:
$P_y=P.\cos\varphi $
Weight in X:
$P_x=P.\sin\varphi$
Tension in Y:
$T_y=T.\sin\alpha$
Tension in X:
$T_x=T.\cos\alpha$
Normal force:
$P_y-N-T_y=0 \implies N = T_y-P_y$
Newton's 2nd law:
$\sum F = m_t.a \implies T_x-F_r-P_x=m_t.a$
Frictional force:
$F_r=\mu _c.N$
work definition:
$W=m_t.a.d$
Total mass:
$m_t=m_d+m_s \implies m_t =30Kg$
I want to get the acceleration of the object and then replacing $a$ I can get the string tension, so (…)
$a={{T_x-F_r-P_x}\over m_t}$
Now, ${T_x-F_r-P_x} = \sum F \implies {T_x-F_r-P_x} = W =m_t.a.d$
But again, If such replacement, the unknown $a$ appears again.
¿How can I solve this?
Best Answer
$\varphi=15°$ ; $m_t=30 kg $; $\mu _c =0,25 $; $ d=130m$;
Newton's 1st law:
$$\sum F_x = 0 \implies T_x-\mu _c m_tg cos \varphi - m_tgsin \varphi= 0$$ $$\sum F_y = 0 \implies T_y-m_tg cos \varphi = 0$$
$$\begin{pmatrix} T_x \\ T_y \end{pmatrix} = \begin{pmatrix} \mu _c m_tg cos \varphi +m_tgsin \varphi \\ m_tg cos \varphi \end{pmatrix} = \begin{pmatrix} 1.23 \\ 0.96 \end{pmatrix} N $$
After you find $T_x$ and $T_y$, you can find the resultant vector which will be in the direction of $\vec d$. The work done then is simply:
$$W=\vec T \bullet \vec d = |T||d|$$