12. QCA block of mass 3.00 kg is pushed up against a wall by aforce P that makes an angle of 0 50.0° with the horizontalas shown in Figure P5.12. The coeffi-cient of static fric tion between the blockand the wall is 0.250. (a) Determine thepossible values for the magnitude of Pthat allow the block to remain station-ary. (b) Describe what happens if |Phas a larger value and what happens if itis smaller. (c) Repeat parts (a) and (b),assuming the force makes an angle of0 13.0° with the horizontal.Figure P5.12

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Asked Oct 31, 2019
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12. QCA block of mass 3.00 kg is pushed up against a wall by a
force P that makes an angle of 0 50.0° with the horizontal
as shown in Figure P5.12. The coeffi-
cient of static fric tion between the block
and the wall is 0.250. (a) Determine the
possible values for the magnitude of P
that allow the block to remain station-
ary. (b) Describe what happens if |P
has a larger value and what happens if it
is smaller. (c) Repeat parts (a) and (b),
assuming the force makes an angle of
0 13.0° with the horizontal.
Figure P5.12
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12. QCA block of mass 3.00 kg is pushed up against a wall by a force P that makes an angle of 0 50.0° with the horizontal as shown in Figure P5.12. The coeffi- cient of static fric tion between the block and the wall is 0.250. (a) Determine the possible values for the magnitude of P that allow the block to remain station- ary. (b) Describe what happens if |P has a larger value and what happens if it is smaller. (c) Repeat parts (a) and (b), assuming the force makes an angle of 0 13.0° with the horizontal. Figure P5.12

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Expert Answer

Step 1

(a)

Write the equilibrium condition for the vertical direction of force.

Write the equilibrium condition for the horizontal direction of force.

The force due to friction is,

mg+fPsin
N Pcose
4Pcos
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mg+fPsin N Pcose 4Pcos

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Step 2

Substitute equation (II) in equation (I), to find maximum force exerted on the block.

mg +д Рсos@— Psin@
mg 3 Psin@-ДРоos0
mg- P(sin@- қоово)
mg
sin @ - д сos0
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mg +д Рсos@— Psin@ mg 3 Psin@-ДРоos0 mg- P(sin@- қоово) mg sin @ - д сos0

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Step 3

Substitute the values,

Similarly the minimum force exerted...

(3.0k/о.8mk?)
Р.
sin(50°)- (0250)cо(50°)
=48.6N
mg
sin@+ дсоsе
(3.0kg)(9.8m/s")
sin (50°) +(0.250)oсов (50°)
-31.7N
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(3.0k/о.8mk?) Р. sin(50°)- (0250)cо(50°) =48.6N mg sin@+ дсоsе (3.0kg)(9.8m/s") sin (50°) +(0.250)oсов (50°) -31.7N

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