A block sits on plane that is inclined to the ground at an a

A block sits on plane that is inclined to the ground at an angle of 30 degrees. A friction force 30N acts a on the block. Which is directed upwards along the plane as shown. Find the resultant force on the block in a direction parallel to the plane. Find an expression for the resultant force on the block in a direction perpendicular to the plane in terms of the forces given. Hence find the normal force F_N on the block.

Solution

2pi/5 :

We add and subtract 2pi....

2pi/5 - 2pi ---> -8pi/5 --> this could be the negative

2pi/5 + 2pi ----> 12pi/5 ---> this could be the positive

Option A

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pi/2 :

pi/2 - 2pi ----> -3pi/2
pi/2 + 2pi ---> 5pi/2

Option C

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-3pi/7 :
Add 2pi to get -3pi/7 + 2pi
11pi/7

This is in quadrant 3

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-664 degrees

Add 360 to make this -304

Add another 360 to make this 56 degrees

Quadrant 1

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6 inches long
16 min

In 16 min, the min hand covers (16/60) * 2pi radians
32pi/60 radians
8pi/15 radians

Distance covered = arc length = L

L = r * theta

L = 6 * (8pi/15)

L = 48pi/15 inches

Option B

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Supp of 5pi/12 :

pi - 5pi/12

(12pi - 5pi) / 12

7pi/12 ----> ANSWER

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Compl of 27 degrees :
90 - 27

63 degrees ----> ANSWER