Use the following situation for the next two problems. Two...

Use the following situation for the next two problems. Two boxes, labeled A and B, are connected by a length of inextensible string that passes over a frictionless and massless pulley as shown in the diagram on the right. Box A is at rest on a ramp with a slope of 30° with respect to horizontal. Box A has inertia of 35 kg, and box B has inertia of 14 kg. 7. (5 points) What is the minimum coefficient of static friction, ?s, between the ramp and the block such that the system remains at rest and in what direction is the static friction force? A) ?s = 0.10, static friction force acts down the ramp. B) ?s = 0.10, static friction force acts up the ramp. C) ?s = 0.12, static friction force acts down the ramp. D) ?s = 0.12, static friction force acts up the ramp. E) ?s = 0.93, static friction force acts down the ramp. 8. (5 points) If ?s is not sufficient to prevent the box from moving, and the coefficient of kinetic friction between the ramp and the block is ?k = 0.092, as the boxes move by some distance, the mechanical energy of the blocks-Earth system is reduced by 55 J. How far did the boxes move? A) 0.9 m B) 1.5 m C) 1.7 m D) 2.0 m E) 3.5 m

Transcript

00:02 Hi, here in this given problem there is an inclined plane, a rough incline plane.

00:07 This is a pulley attached with it at the top, a block a at the inclined plane, a mass less string passing over this massless pulley, and this is another block, block b attached here.

00:38 Mass means inertia of the block, inertia of the block means mass of the block that is for a that is given as 35 kilogram for b this is just 14 kilograms so being heavy the block a will be having a tendency to slide down like this angle is 30 degree so as the block is having a tendency to slide down a force of friction will be acting and upward static friction now the other forces are for b its weight m b g acting downward tension t upward weight of block a that will also be acting vertically downward m a g but its component will be if this angle is 30 degree this will also be 30 degrees so the component is component of the force weight along the incline that is m, a, g, sine 30 degree, normal reaction n, tension in the string t on both the segments of this string.

02:02 In this seventh question here, we have to find coefficient of static friction between the inclined plane and the block provided the system is at rest.

02:17 So as the blocks are at rest means there will be no acceleration in the blocks.

02:30 So using free -body diagram of block b, t will be equal to m -b -g as there is no motion, no net motion in the block.

02:52 So, there should be no net force acting over there.

02:56 So t is equal to m bg.

02:58 Then using fdr of block a here two forces are upward, one forces downward.

03:08 So this is t plus f s force of friction that is equal to m a g, sine 30 degree.

03:24 Or we can say for tension t we have found it that is m bg.

03:30 M b into g and force of friction is given as mu s times the normal reaction and normal reaction is equal to m a g cos theta is equal to m a g sine theta or 30 whatever we can say so now putting all these values all the known values here for mu s this is missing we have to find it m .a, that is 35, g 9 .8, cost 30 degree is equal to m .a, again 35 into g 9 .8 into sine 30 degree and minus mbg 14 into 9 .8.

04:17 So here it comes out to be equal to in right -hand side, 171 .5 minus 137.

04:26 So, finally, mu s will be given by this remaining will be 34 .3 divided by this, which comes out to be equal to 297 .05.

04:42 So finally, coefficient of static friction here, that is 0 .115, or we can say in two decimal places, this is 0 .12...

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