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A box with a mass of 2 $\mathrm{kg}$ is placed on an inclined plane that makes a $30^{\circ}$ angle with the horizontal. What must the coefficient of static fraction $\left(\mu_{\mathrm{s}}\right)$ between the box and the inclined be in order for the box to be at rest?(A) 0.5(B) 0.58(C) 0.87(D) 1

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Chapter 13

Practice Test 3

Section 1

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Hi in the given problem here, this is an inclined plane which is making an angle of 30°.. With the horizontal. There is a block kept over this inclined plane mas of the box. It's given as M is equal to two kg force of friction acting on this block will be given by F. Is equal to new. Not times we are. This newest is the coefficient of static friction. So this is new as times the normal reaction. Now, if this is the weight of this box acting vertically downward, It will be having two components 1 normal to this inclined plane. If this angle is 30°,, this will also, with 30° of this component normal to the inclined plane will be MG cost 30° and the component of the weight along the inclined plane which will be having a tendency to bring it down. That is MG signed 30 degree. Now the normal reaction of the inclined plane on the block will be equal to this. Mg cost 30 degree. So four the box to be addressed. The force of friction will be acting here along the inclined plane in upward direction because the tendency of the box is to move downward and we know for so friction always X oppose it to the direction of motion. So for the box to be addressed, the downward force, MG sine 30 degree should be equal to upward force means force of friction means this is you know, times the normal reaction or we can say this is new, not times MG Cost 30°.. So finally canceling this MG. An expression for this coefficient of static friction comes out to be signed 30°,, divided by cost 30° means this is 10 30 degree, whose values one by route three. Or we can say this is zero point 58 Hence here we can see about option B is correct here. Thank you.

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