Two wedges each of mass 600 g are placed nex | vedclass

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(A) Let $m = 0.6 \ kg$ be the mass of each wedge. The angle of the wedge is $	heta = 45^{\circ}$.For the cube of mass $M$,the normal force $N$ exerte

Vedclass · Accepted Answer

$0.8$

Vedclass · Answer

(A) Let $m = 0.6 \ kg$ be the mass of each wedge. The angle of the wedge is $	heta = 45^{\circ}$.For the cube of mass $M$,the normal force $N$ exerted by each wedge on the cube is given by $2N \cos(45^{\circ}) = Mg$,so $N = \frac{Mg}{2 \cos(45^{\circ})} = \frac{Mg}{\sqrt{2}}$.By Newton's third law,the cube exerts a normal force $N$ on the wedge at an angle of $45^{\circ}$ to the horizontal.The horizontal component of this force is $N \cos(45^{\circ}) = \frac{Mg}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{Mg}{2}$.This horizontal force tends to push the wedge outward.The maximum static frictional force that opposes this motion is $f_{max} = \mu N_{total}$,where $N_{total}$ is the total normal force on the ground.The vertical forces on the wedge are its weight $mg$,the vertical component of the force from the cube $N \sin(45^{\circ}) = \frac{Mg}{2}$,and the normal force from the ground $N_g$.So,$N_g = mg + \frac{Mg}{2}$.The condition for equilibrium is $f_{max} \ge \frac{Mg}{2}$.Thus,$\mu (mg + \frac{Mg}{2}) \ge \frac{Mg}{2}$.Substituting $\mu = 0.4$ and $m = 0.6 \ kg$:$0.4 (0.6g + 0.5Mg) \ge 0.5Mg$.$0.24g + 0.2Mg \ge 0.5Mg$.$0.24g \ge 0.3Mg$.$M \le \frac{0.24}{0.3} = 0.8 \ kg$.Therefore,the largest mass $M$ is $0.8 \ kg$.

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