A parcel of mass 4kg lies on a rough slope inclined at 300 to the horizontal. The parcel is held in - PowerPoint PPT Presentation

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A parcel of mass 4kg lies on a rough slope inclined at 300 to the horizontal. The parcel is held in

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A parcel of mass 4kg lies on a rough slope inclined at 300 to the horizontal. The parcel is held in equilibrium by a force of 45 N acting at 500 to the plane. ... – PowerPoint PPT presentation

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Title: A parcel of mass 4kg lies on a rough slope inclined at 300 to the horizontal. The parcel is held in


1
45 N
4kg
500
300
A parcel of mass 4kg lies on a rough slope
inclined at 300 to the horizontal. The parcel is
held in equilibrium by a force of 45 N acting at
500 to the plane. The parcel is in equilibrium an
on the point of moving up the plane. Find a. the
normal reaction to the plane b. the coeficient of
friction.
2
45 N
4kg
500
4g cos30
45sin50
45cos50
300
4g
300
4g sin30
Normal reaction 45sin50 4gcos30
? 34.47 33.95
? 68.42 Newtons
3
F
45 N
4kg
The particle is in equilibrium on the point of
moving up the slope
NB Friction acts against direction of movement
500
4g cos30
45sin50
45cos50
300
4g
300
Normal reaction 68.42 Newtons Friction force
F 68.42?
4g sin30
45cos50 4gsin30 68.42? 28.93 19.6
68.42? 68.42? 9.33 ? 0.136
4
M1 Jan 06
A parcel of weight 10 N lies on a rough plane
inclined at an angle of 30? to the horizontal.
A horizontal force of magnitude P Newtons acts on
the parcel, as shown in Figure 2. The parcel is
in equilibrium and on the point of slipping up
the plane. The normal reaction of the plane on
the parcel is 18 N. The coefficient of friction
between the parcel and the plane is ?.
Find (a) the value of P,
(4) (b) the value of ?.
(5) The horizontal force is removed. (c) Determi
ne whether or not the parcel moves.
5
Pcos30
F
Psin30
Friction force F ? ?Pcos30
10cos30
10N
10sin30
The normal reaction of the plane is 18 ? 18
10cos30 Psin30 ? 18 8.66 0.5P ? P 18.7 N
Since the block is on the point of moving Pcos30
10sin30 ?10cos30 18.7 (?sin30) 16.2 5
8.6? 9.4? ? 18? 11.2 ? ? 0.62
6
250
Force X to push up the slope so that the particle
is on the point of moving
X
F
Xcos25
F is the force created by friction
2kg
450
? 0.2
2g cos20
Xsin25
NB Friction acts against direction of movement
200
2g
F ?2gcos20 ? Xcos25
200
2g sin20
Xsin25 2gsin20 F ? Xcos20 2gsin20 -
F Xsin25 2gsin20 ?2gcos20 ?Xcos25 0.42X
6.70 3.68 0.18X 0.24X 10.38 ? X 43.3
Newtons
7
250
X
F
Force X to stop the particle sliding down the
slope
Xcos25
F is the force created by friction
2kg
450
? 0.2
2g cos20
Xsin25
200
NB Friction acts against direction of movement
2g
F ?2gcos20 ? Xcos25
200
2g sin20
Xsin25 F 2gsin20 Xsin25 2gsin20 (?2gcos20
?Xcos25) 0.42X 6.70 - 3.68 - 0.18X 0.6X
3.02 ? X 5 Newtons
8
M1 Jan 06
A parcel of weight 10 N lies on a rough plane
inclined at an angle of 30? to the horizontal.
A horizontal force of magnitude P Newtons acts on
the parcel, as shown in Figure 2. The parcel is
in equilibrium and on the point of slipping up
the plane. The normal reaction of the plane on
the parcel is 18 N. The coefficient of friction
between the parcel and the plane is ?.
Find (a) the value of P,
(4) (b) the value of ?.
(5) The horizontal force is removed. (c) Determi
ne whether or not the parcel moves.
9
Pcos30
F
Psin30
Friction force F ? ?Pcos30
10cos30
10N
10sin30
The normal reaction of the plane is 18 ? 18
10cos30 Psin30 ? 18 8.66 0.5P ? P 18.7 N
Since the block is on the point of moving Pcos30
10sin30 ?10cos30 18.7 (?sin30) 16.2 5
8.6? 9.4? ? 18? 11.2 ? ? 0.62
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