5. Forces and equilibrium
From Mechanics
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Resolving vertically gives; | Resolving vertically gives; | ||
| - | <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 \ \text{N}</math> |
Now solving the equations by substituting | Now solving the equations by substituting | ||
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<math>\begin{align} | <math>\begin{align} | ||
| - | & {{T}_{1}}\sin 60{}^\circ =58\textrm{.}8 \\ | + | & {{T}_{1}}\sin 60{}^\circ =58\textrm{.}8 \ \text{N}\\ |
& {{T}_{1}}=\frac{58\textrm{.}8}{\sin 60{}^\circ }=67\textrm{.}9\text{ N (to 3 sf)} \\ | & {{T}_{1}}=\frac{58\textrm{.}8}{\sin 60{}^\circ }=67\textrm{.}9\text{ N (to 3 sf)} \\ | ||
\end{align}</math> | \end{align}</math> | ||
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'''[[Example 5.4]]''' | '''[[Example 5.4]]''' | ||
| - | A child, of mass | + | A child, of mass 30 kg, 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 <math>{{40}^{\circ }}</math> with the horizontal. Find the magnitude of the friction force on the child and the coefficient of friction. |
'''Solution''' | '''Solution''' | ||
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Resolving perpendicular to the slope gives: | Resolving perpendicular to the slope gives: | ||
| - | <math>R=294\cos 40{}^\circ </math> | + | <math>R=294\cos 40{}^\circ \ \text{N}</math> |
As the child is sliding | As the child is sliding | ||
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| - | Note – Angle of Friction | + | |
| + | '''[[Note – Angle of Friction]]''' | ||
[[Image:AngleFriction.gif]] | [[Image:AngleFriction.gif]] | ||
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Resolving horizontally: | Resolving horizontally: | ||
| - | <math>F=T\cos 20{}^\circ </math> | + | <math>F=T\cos 20{}^\circ \ \text{N}</math> |
Resolving vertically: | Resolving vertically: | ||
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or | or | ||
| - | <math>R=1960-T\sin 20{}^\circ </math> | + | <math>R=1960-T\sin 20{}^\circ \ \text{N}</math> |
As the crate is sliding we can use | As the crate is sliding we can use | ||
Revision as of 11:56, 11 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.
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.
Solution
The diagram shows the forces acting on the object.
Resolving horizontally or using the horizontal components of the forces:
=T2cos60T1=T2
Resolving vertically gives;
+T2sin60=2940 N
Now solving the equations by substituting
gives:
+T2sin60=29402T1sin60=2940T1=29402sin60=1700 N (to 3sf)
And also
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
The diagram shows the forces acting on the particle.
Resolving vertically:
=58.8 NT1=58.8sin60=67.9 N (to 3 sf)
Resolving horizontally:
=T2T2=58.8sin60cos60=33.9 N (to 3sf)
A lorry of mass 5000 kg drives up a slope inclined at
Solution
Model the lorry as a particle.
The diagram shows the forces acting on the lorry.
Resolving perpendicular to the slope gives:
=48800 N (to 3 sf)
Resolving parallel to the slope gives:
=4270 N (to 3sf)
A child, of mass 30 kg, 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
Solution
Model the child as a particle.
The diagram shows the forces acting on the child.
Resolving parallel to the slope gives.
=189 N (to 3sf)
Resolving perpendicular to the slope gives:
N
As the child is sliding
R
=
294cos40=294sin40294cos40=tan40=0.840 (to 3 sf)
If a particle of mass 
Then using
R
mgcos
sincostan
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
The diagram shows the forces acting on the crate.
Resolving horizontally:
N
Resolving vertically:
=1960
or
N
As the crate is sliding we can use
R
Using this equation with the horizontal and vertical equations gives:
=0.4
1960−Tsin20
Tcos20=784−T0.4sin20Tcos20+T0.4sin20=784Tcos20+0.4sin20=784T=784cos20+0.4sin20=728 N (to 3sf)
