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5. Forces and equilibrium

From Mechanics

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{{Selected tab|[[5. Forces and Equilibrium|Theory]]}}
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5. Forces and Equilibrium
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== '''Key Points''' ==
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Key Points
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If the resultant of the forces acting on a particle is zero we say that these forces are in equilibrium.
If the resultant of the forces acting on a particle is zero we say that these forces are in equilibrium.
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'''[[Example 5.1]]'''
'''[[Example 5.1]]'''
- 
- 
-
[[Image:ex5.5fig1.gif]]
 
- 
The diagram shows an object, of mass 300 kg, that is at rest and is supported by two cables. Find the tension in each cable.
The diagram shows an object, of mass 300 kg, that is at rest and is supported by two cables. Find the tension in each cable.
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'''Solution'''
'''Solution'''
- 
- 
- 
The diagram shows the forces
The diagram shows the forces
acting on the object.
acting on the object.
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- 
- 
- 
- 
- 
Resolving horizontally or using the horizontal components of the forces:
Resolving horizontally or using the horizontal components of the forces:
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<math>\begin{align}
<math>\begin{align}
& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}}\cos 60{}^\circ \\ & {{T}_{1}}={{T}_{2}}
& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}}\cos 60{}^\circ \\ & {{T}_{1}}={{T}_{2}}
\end{align}</math>
\end{align}</math>
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Resolving vertically gives;
Resolving vertically gives;
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<math>{{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940</math>
<math>{{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940</math>
- 
Now solving the equations by substituting
Now solving the equations by substituting
<math>{{T}_{1}}={{T}_{2}}</math>
<math>{{T}_{1}}={{T}_{2}}</math>
 +
gives:
gives:
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<math>\begin{align}
<math>\begin{align}
& {{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940 \\
& {{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940 \\
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& {{T}_{1}}=\frac{2940}{2\sin 60{}^\circ }=1700\text{ N (to 3sf)} \\
& {{T}_{1}}=\frac{2940}{2\sin 60{}^\circ }=1700\text{ N (to 3sf)} \\
\end{align}</math>
\end{align}</math>
- 
And also
And also
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<math>{{T}_{2}}=1700\text{ N (to 3 sf)}</math>
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<math>{{T}_{2}}=1700\text{ N (to 3 sf)}</math>.
-
.
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'''[[Example 5.2]]'''
'''[[Example 5.2]]'''
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strings as shown in the diagram. Note that one
strings as shown in the diagram. Note that one
string is horizontal. Find the tension in each string.
string is horizontal. Find the tension in each string.
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'''Solution'''
'''Solution'''
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Resolving vertically:
Resolving vertically:
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<math>\begin{align}
<math>\begin{align}
& {{T}_{1}}\sin 60{}^\circ =58.8 \\
& {{T}_{1}}\sin 60{}^\circ =58.8 \\
& {{T}_{1}}=\frac{58.8}{\sin 60{}^\circ }=67.9\text{ N (to 3 sf)} \\
& {{T}_{1}}=\frac{58.8}{\sin 60{}^\circ }=67.9\text{ N (to 3 sf)} \\
\end{align}</math>
\end{align}</math>
- 
Resolving horizontally:
Resolving horizontally:
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<math>\begin{align}
<math>\begin{align}
& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}} \\
& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}} \\
& {{T}_{2}}=\frac{58.8}{\sin 60{}^\circ }\cos 60{}^\circ =33.9\text{ N (to 3sf)} \\
& {{T}_{2}}=\frac{58.8}{\sin 60{}^\circ }\cos 60{}^\circ =33.9\text{ N (to 3sf)} \\
\end{align}</math>
\end{align}</math>
- 

Revision as of 10:04, 9 February 2010

       Theory          Exercises      


Key Points

If the resultant of the forces acting on a particle is zero we say that these forces are in equilibrium.


Example 5.1

The diagram shows an object, of mass 300 kg, that is at rest and is supported by two cables. Find the tension in each cable.

Image:TF5.1.GIF

Solution

The diagram shows the forces acting on the object.

Resolving horizontally or using the horizontal components of the forces:

T1cos60=T2cos60T1=T2

Resolving vertically gives;

T1sin60+T2sin60=2940

Now solving the equations by substituting T1=T2

gives:

T1sin60+T2sin60=29402T1sin60=2940T1=29402sin60=1700 N (to 3sf)

And also T2=1700 N (to 3 sf).


Example 5.2

Image:ex5.2fig1.gif

A particle of mass 6 kg is suspended by two strings as shown in the diagram. Note that one string is horizontal. Find the tension in each string.

Solution

Image:ex5.2fig2.gif

The diagram shows the forces acting on the particle.

Resolving vertically:

T1sin60=588T1=588sin60=679 N (to 3 sf)

Resolving horizontally:

T1cos60=T2T2=588sin60cos60=339 N (to 3sf)


Example 5.3

A lorry of mass 5000 kg drives up a slope inclined at 5 to the horizontal. The lorry moves in a straight line and at a constant speed. Assume that no resistance forces act on the lorry. Find the magnitude of the normal reaction force and force that acts on the lorry in its direction of motion.

Solution

Image:ex5.3fig1.gif


Model the lorry as a particle.

The diagram shows the forces acting on the lorry.

Resolving perpendicular to the slope gives:


R=49000cos5=48800 N (to 3 sf)


Resolving parallel to the slope gives:


P=49000sin5=4270 N (to 3sf)


Example 5.4

A child, of mass 30kg, slides down a slide at a constant speed. Assume that there is no air resistance acting on the child. The slide makes an angle of 40 with the horizontal. Find the magnitude of the friction force on the child and the coefficient of friction.

Solution

Model the child as a particle.


Image:ex5.4fig1.gif

The diagram shows the forces acting on the child.

Resolving parallel to the slope gives.


F=294sin40=189 N (to 3sf)


Resolving perpendicular to the slope gives:


R=294cos40


As the child is sliding F=R so that we can determine .


294sin40=294cos40=294sin40294cos40=tan40=0840 (to 3 sf)



Note – Angle of Friction


Image:AngleFriction.gif

If a particle of mass m is at rest on a slope at an angle above the horizontal, then :


F=mgsin


R=mgcos


Then using FR gives:


mgsinmgcossincostan



Example 5.5


Image:ex5.5fig1.gif


A crate of mass 200 kg is on a horizontal surface. The coefficient of friction between the crate and the surface is 0.4. The crate is pulled by a rope, as shown in the diagram, so that the crate moves at a constant speed in a straight line. Find the tension in the rope.


Solution


Image:ex5.5fig2.gif


The diagram shows the forces acting on the crate.

Resolving horizontally:


F=Tcos20


Resolving vertically:


R+Tsin20=1960

or

R=1960−Tsin20


As the crate is sliding we can use F=R, which gives:


F=04R


Using this equation with the horizontal and vertical equations gives:


Tcos20=041960−Tsin20Tcos20=784−T04sin20Tcos20+T04sin20=784Tcos20+04sin20=784T=784cos20+04sin20=728 N (to 3sf)

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