Answer to Question #213657 in Mechanics | Relativity for anuj

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

For a horizontally applied force:

(a) The force is

$F=\mu_s mg \\ F = 0.23 \times 4 \times 9.8 = 9.016 \;N$

(b) The force is

$F= \mu_k mg \\ F=0.18 \times 4 \times 9.8 = 7.056 \;N$

For a force applied at 25°:

(a) The force is

$O_x: F cosθ -\mu_k N=0 \\ O_y: F sin θ -mg+N=0 \\ F = \frac{\mu_s mg}{cos25° + \mu_s sin25°} \\ F = \frac{9.016}{0.9063 + 0.23 \times 0.4226} \\ = \frac{9.016}{1.0034}= 8.985 \;N$

(b) The force is

$F = \frac{\mu_k mg}{cos25° + \mu_k sin25°} \\ F = \frac{7.056}{0.9063 + 0.18 \times 0.4226} \\ = \frac{7.056}{0.9823}= 7.183 \;N$