Question

Block A has a mass of $30\text{kg}$ and block B a mass of $15\text{kg}$. The coefficients of friction between all surfaces of contact are ${\mu }_s=0.15$ and ${\mu }_k=0.10$. Knowing that $\theta ={30}^{\circ }$ and that the magnitude of the force $F$ applied to block A is $250\text{N}$, determine the accelaration of block A.

[Take $g=10{\text{m/s}}^2$]

Answer 1

FBD of block A and B are given below. Assume friction $f$ acts between the blocks and $f^{\prime }$ is the friction from the ground on block B.

The blocks move with same acceleration $a$ relative to each other [due to string constraint]. Hence, friction will be kinetic.


For A:
$m_{A}g\sin \theta +F-T-f=m_{A}a\dots \left(1\right)$
where $f={\mu }_{k}N={\mu }_k{m}_{A}g\cos \theta =30\cos \theta$
For B:
$T-m_{B}g\sin \theta -f-f^{{}^{\prime }}=m_{B}a\dots \left(2\right)$
where $f^{\prime }={\mu }_k{N}^{\prime }={\mu }_k(m_{A}g\cos \theta +m_{B}g\cos \theta )$
$=45\cos \theta$

Adding (1) and (2),
$(m_A-m_B)g\sin \theta +F-f-f^{\prime }=(m_A+m_B)a$
$\Rightarrow 15g\sin \theta +250-105\cos \theta =45a$
[$\theta ={30}^{\circ }$]
or $a=5.2{\text{m/s}}^2$