It is question about forces and equilibrium ?

A block of mass 20 kg is at rest on a plane inclined at 10◦ to the horizontal. A force acts on the block
parallel to a line of greatest slope of the plane. The coefficient of friction between the block and the
plane is 0.32. Find the least magnitude of the force necessary to move the block,
(i) given that the force acts up the plane,
(ii) given instead that the force acts down the plane.

2 Answers
Tom M.
May 13, 2018

Up the slope - #96.9N#
Down the slope - #23.2N#

Explanation:

Draw a forces diagram for these situations.

enter image source here

X is the force we exert up the slope.

Begin by resolving perpendicular to the slope using Newton's First Law (it is not accelerating into the slope).

#20gcos10^@=R#

To find friction:

Using #F_(max)=muR#
#F_max=0.32xx20gcos10#
#=6.4gcos10#

Resolve up the slope using Newton's Second Law.
(I'll let the acceleration be #a# for now).

#X-6.4gcos10-20gsin10=20a#
#X=6.4cos10+20gsin10+20a#

Since we're looking for the minimal force to move the block, we can let #a=0.#
#X>6.4gcos10+20gsin10#
#X>96.9N#

We take a similar scenario for when the force is acting down the slope.

enter image source here

Again, from Newton's First Law,

#R=20gcos10#, which also means #F_max=6.4gcos10#

The difference comes when we resolve parallel to the slope using Newton's Second Law.
Let's resolve down the slope.

#X+20gsin10-6.4gcos10=20a#
#X=6.4gcos10-20gsin10+20a#
let #a=0#
#X>6.4gcos10-20gsin10#
#X>27.8N#

Somebody N.
May 13, 2018

See below.

Explanation:

i).

enter image source here

Resolving perpendicular to the plane:

#R=20gcos(10) \ \ \ \ \ \[1]#

Parallel to the plane:

#P=Fr+20gsin(10) \ \ \ \ \[2]#

Limiting equilibrium:

#Fr=muR \ \ \ \ \[3]#

#[1] "into" [3]#

#Fr=mu(20gcos(10)) \ \ \ \ [4]#

#[4] "into" [2]#

#P=mu20gcos(10)+20gsin(10)#

Let #g=9.8#

#mu =0.32#

Plugging in these values:

#P=0.32*20(9.8)cos(10)+20(9.8)sin(10)=95.80N#

This is for limiting equilibrium, so:

#P>95.80N#

ii).

enter image source here

Resolving perpendicular to the plane:

#R=20gcos(10) \ \ \ \[1]#

Parallel to the plane:

#P+Fr=20gsin(10) \ \ \ \[2]#

Limiting equilibrium:

#Fr=muR \ \ \ \[3]#

#[1] "into" [3]#

#Fr=mu20gcos(10) \ \ \ \[4]#

#[4] "into" [2]#

#P=20gsin(10)-mu20gcos(10)#

Plugging in values:

#P=20(9.8)sin(10)-0.32*20(9.8)cos(10)=-27.73N#

This is for limiting equilibrium, so:

#P>27.73N#

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