91Ó°ÊÓ

M =11 kg

The coefficient of static friction between block and table is${\mathrm{μ}}_s=0.52$

Angle that force acting from the horizontal$\mathrm{θ}=37°$

Frictional force is given by the product of coefficient of friction and the normal reaction. Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass. Using these concepts, the problem can be solved.

(a)

We know $F_N=mg$

$$\begin{gathered} f_{s,max}={\mathrm{μ}}_{s}mg \\ =\left(0.52\right)\left(11kg\right)\left(9.8m/s^2\right) \\ =56N \end{gathered}$$

Consequently, the horizontal force F needed to initiate motion must be slightly more than 56 N

(b)

Analyzing vertical forces when F is at nonzero yields

$F_N+F\sin \mathrm{θ}=mg;f_{s,max}={\mathrm{μ}}_s\left(mg-F\sin \mathrm{θ}\right)$

The horizontal component of F needed to initiate motion must be slightly more than this, so

$$\begin{gathered} F\cos \mathrm{θ}={\mathrm{μ}}_s\left(mg-F\sin \mathrm{θ}\right) \\ ⇒F=\frac{{\mathrm{μ}}_{s}mg}{\cos \mathrm{θ}+{\mathrm{μ}}_s\sin \mathrm{θ}} \end{gathered}$$

which yields F = 59 N when θ = 60°

(c)

At θ = -60°

$$\begin{gathered} \mathrm{F}=\frac{\left(0.52\right)\left(11\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\cos \left(-60°\right)+\left(0.52\right)\sin \left(-60°\right)} \\ =1,1×{10}^3\mathrm{N} \end{gathered}$$