Question

Block $A$ and $B$ shown in the figure are connected with a bar of negligible weight. $A$ and $B$ each has mass $170\text{kg}$, the coefficient of friction between $A$ and the plane is $0.2$ and that between $B$ and the plane is $0.4$ ($g=10{\text{ms}}^{-2}$)
What is the total force of friction between the blocks and the plane?

[Hint: ${\tan }^{-1}\left(\frac{8}{15}\right)={28}^{\circ },\cos {28}^{\circ }=0.883,\sin {28}^{\circ }=0.469$]

• A. $901\text{N}$
• B. $760\text{N}$
• C. $600\text{N}$
• D. $300\text{N}$

Answer 1

The correct option is A $901\text{N}$

$\tan \theta =\frac{8}{15}$
$\therefore \theta ={\tan }^{-1}\left(\frac{8}{15}\right)={28}^{\circ }$
$(f_A{)}_{\text{max}}=0.2\times 170\times 10\times \cos {28}^{\circ }$
$=300.22\text{N}$
$(f_B{)}_{\text{max}}=0.4\times 170\times 10\times \cos {28}^{\circ }$
$=600.44\text{N}$
Now,
$(m_A+m_B)g\sin \theta =\left(340\right)\left(10\right)\sin {28}^{\circ }=1594.6\text{N}$
Since this is greater than $(f_A{)}_{\text{max}}+(f_B{)}_{\text{max}}$, therefore blocks slides downward and maximum force of friction will act on both surfaces
$\therefore f_{\text{total}}=(f_A{)}_{\text{max}}+(f_B{)}_{\text{max}}$
$=300.22+600.44=900.66\approx 901\text{N}$