In the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 106.6N. The coefficient of kinetic friction between the lower block and the surface is 0.257. The coefficient of kinetic friction between the lower block and the upper block is also 0.257. What is the acceleration of the lower block, if the mass of the lower block is 4.82kg and the mass of the upper block is 2.05kg?
/-------m1
|O
\--------m2--->F
The lines represent the string and m1 sits on top of m2 which is on the floor.
+33657 Normal force of surface on lower block = (2.05+4.82)*9.81N,
hence horizontal friction force = 0.257*(2.05+4.82)*9.81N.
Vertical force of upper block on lower block = 2.05*9.81N,
hence horizontal friction force = 0.257*2.05*9.81N.
Newton's 2nd law on lower block:
4.82*a = 106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81
a = (106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81)/4.82
$${\mathtt{a}} = {\frac{\left({\mathtt{106.6}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}\left({\mathtt{2.05}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.82}}\right){\mathtt{\,\times\,}}{\mathtt{9.81}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}{\mathtt{2.05}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}{{\mathtt{4.82}}}} \Rightarrow {\mathtt{a}} = {\mathtt{17.450\: \!448\: \!879\: \!668\: \!049\: \!8}}$$
acceleration ≈ 17.45 m/s2
.
+33657 Normal force of surface on lower block = (2.05+4.82)*9.81N,
hence horizontal friction force = 0.257*(2.05+4.82)*9.81N.
Vertical force of upper block on lower block = 2.05*9.81N,
hence horizontal friction force = 0.257*2.05*9.81N.
Newton's 2nd law on lower block:
4.82*a = 106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81
a = (106.6 - 0.257*(2.05+4.82)*9.81 - 0.257*2.05*9.81)/4.82
$${\mathtt{a}} = {\frac{\left({\mathtt{106.6}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}\left({\mathtt{2.05}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.82}}\right){\mathtt{\,\times\,}}{\mathtt{9.81}}{\mathtt{\,-\,}}{\mathtt{0.257}}{\mathtt{\,\times\,}}{\mathtt{2.05}}{\mathtt{\,\times\,}}{\mathtt{9.81}}\right)}{{\mathtt{4.82}}}} \Rightarrow {\mathtt{a}} = {\mathtt{17.450\: \!448\: \!879\: \!668\: \!049\: \!8}}$$
acceleration ≈ 17.45 m/s2
.
