Question

An object of mass 10 kg is to be kept at rest on an inclined plane making an angle of ${37}^0$ to the horizontal by applying a force F along the plane upwards as shown in Fig. The coefficient of static friction between the object and the plane is 0.2. Find the magnitude of force F.
[Take g= 10 $m{s}^{-2}$]

Answer 1

$mgsin{37}^0=10\times 10\times 3/5=60N$
$f_l=\mu =0.2mgcos{37}^0$
$=0.2\times 10\times 10\times 4/5$
=16N
Friction force alone is unable to balance $mgsin{37}^0$. Hence, F is required for balancing. Here two cases arise.
Case I:
If the block has tendency to slide down, friction on block will act upwards.
F + 16 = 60
$RightarrowF=44N$
This is the minimum force required for balancing.
Case II: If the block has on tendency to slide up, friction on block will act downwards.
F = 60 + 16$\Rightarrow$ F= 76 N
This is the maximum force required for balancing.
Therefore, for balancing $44N\le F\le 16N$