Question

A block 4kg in mass rests on a horizontal plane. What value of horizontal force is required to (a) move the block, (b) maintain the motion of the block once it is moved? The coefficients of static and kinetic friction are 0.23 and 0.18, respectively. What will the force be, if the force is instead applied 25° above the horizontal?

Answer 1

Solution.

$m=4kg;$

$\mu_s=0.23;$

$\mu_k=0.18;$

$a) F=\mu_smg;$ $F=0.23\sdot4kg\sdot9.8N/kg=9N;$

$b) F=\mu_kmg; F=0.18\sdot4kg\sdot9.8N/kg=7N;$

$\theta=25^o;$

$Ox:Fcos\theta-\mu_kN=0;$

$Oy: Fsin\theta+N-mg=0;$

$F=\dfrac{\mu_kmg}{cos\theta+\mu_ksin\theta};$

$F=\dfrac{0.18\sdot4kg\sdot9.8N/kg}{0.9063+0.18\sdot0.4226}=7.2N;$

$F=\dfrac{\mu_smg}{cos\theta+\mu_ssin\theta};$

$F=\dfrac{0.23\sdot4kg\sdot9.8N/kg}{0.9063+0.23\sdot0.4226}=8.99N;$

Answer: $a)9N;$

$b)7N;$

$8.98N; 7.2N$