5. Forces and equilibrium

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Revision as of 15:47, 17 September 2009

       Theory          Exercises      


5. Forces and Equilibrium

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


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.

Image:TF5.1.GIF

Solution



The diagram shows the forces acting on the object.




Resolving horizontally or using the horizontal components of the forces:


\displaystyle \begin{align} & {{T}_{1}}\cos 60{}^\circ ={{T}_{2}}\cos 60{}^\circ \\ & {{T}_{1}}={{T}_{2}} \end{align}


Resolving vertically gives;


\displaystyle {{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940


Now solving the equations by substituting \displaystyle {{T}_{1}}={{T}_{2}} gives:


\displaystyle \begin{align} & {{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940 \\ & 2{{T}_{1}}\sin 60{}^\circ =2940 \\ & {{T}_{1}}=\frac{2940}{2\sin 60{}^\circ }=1700\text{ N (to 3sf)} \\ \end{align}


And also \displaystyle {{T}_{2}}=1700\text{ 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:


\displaystyle \begin{align} & {{T}_{1}}\sin 60{}^\circ =58.8 \\ & {{T}_{1}}=\frac{58.8}{\sin 60{}^\circ }=67.9\text{ N (to 3 sf)} \\ \end{align}


Resolving horizontally:


\displaystyle \begin{align} & {{T}_{1}}\cos 60{}^\circ ={{T}_{2}} \\ & {{T}_{2}}=\frac{58.8}{\sin 60{}^\circ }\cos 60{}^\circ =33.9\text{ N (to 3sf)} \\ \end{align}


Example 5.3

A lorry of mass 5000 kg drives up a slope inclined at \displaystyle {{5}^{\circ }} 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:


\displaystyle R=49000\cos 5{}^\circ =48800\text{ N (to 3 sf)}


Resolving parallel to the slope gives:


\displaystyle P=49000\sin 5{}^\circ =4270\text{ 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 \displaystyle {{40}^{\circ }} 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.


\displaystyle F=294\sin 40{}^\circ =189\text{ N (to 3sf)}


Resolving perpendicular to the slope gives:


\displaystyle R=294\cos 40{}^\circ


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


\displaystyle \begin{align} & 294\sin 40{}^\circ =\mu \times 294\cos 40{}^\circ \\ & \mu =\frac{294\sin 40{}^\circ }{294\cos 40{}^\circ }=\tan 40{}^\circ =0.840\text{ (to 3 sf)} \\ \end{align}



Note – Angle of Friction


Image:AngleFriction.gif

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


\displaystyle F=mg\sin \alpha


\displaystyle R=mg\cos \alpha


Then using \displaystyle F\le \mu R gives:


\displaystyle \begin{align} & mg\sin \alpha \le \mu mg\cos \alpha \\ & \mu \ge \frac{\sin \alpha }{\cos \alpha } \\ & \mu \ge \tan \alpha \\ \end{align}



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:


\displaystyle F=T\cos 20{}^\circ


Resolving vertically:


\displaystyle R+T\sin 20{}^\circ =1960

or

\displaystyle R=1960-T\sin 20{}^\circ


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


\displaystyle F=0.4R


Using this equation with the horizontal and vertical equations gives:


\displaystyle \begin{align} & T\cos 20{}^\circ =0.4\left( 1960-T\sin 20{}^\circ \right) \\ & T\cos 20{}^\circ =784-T\times 0.4\sin 20{}^\circ \\ & T\cos 20{}^\circ +T\times 0.4\sin 20{}^\circ =784 \\ & T\left( \cos 20{}^\circ +0.4\sin 20{}^\circ \right)=784 \\ & T=\frac{784}{\cos 20{}^\circ +0.4\sin 20{}^\circ }=728\text{ N (to 3sf)} \\ \end{align}