Question

1.) what is friction?
2.) explain friction as the component of contact force?
3.) explain why static force is a self adjusting force?
4.) show resultant force exerted by surface on block?
5.) relation between angle of friction and coefficient of friction?


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Answer 1

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

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1.) what is friction?

A. Whenever a body moves or tends to move over the surface of another body a force comes into play which acts parallel to the surface of contact and opposes the relative motion. This opposing force is called fraction.

◉ EXAMPLE:-

Consider a wooden block placed on a horizontal surface. Give it a gentle Push. The block slides through a small distance and then comes back to rest according to Newton's second law. A retarding force must be acting on the block. this opposing force is called friction.

✷The force of friction always acts tangential to the surfaces in contact and in a direction opposite to the direction of (relative) motion of the body.

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2.) explain friction as the component of contact force?

A. When two bodies are placed in contact, attractive forces act between the particles at the surface of contact. As a result, each body exerts a contact force on the other.

These mutual contact forces are equal and opposite, obeying Newton's third law. The component of the contact force F normal to the contact force is called normal force or normal reaction N. The component parallel to the contact force is called friction.

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3.) explain why static force is a self adjusting force?

A. "The force of friction which comes into play between two bodies before one body actually starts moving over the other is called static friction (fs)".

Static friction opposes the impending motion that is, The motion that would take place (but does not take place) under the applied force, if friction were absent.

"The maximum force of static friction(fs max) which comes into play when a body just starts moving over the surface of another body is called limiting friction.

$: \longrightarrow{ \bold {\tt {\purple{clearly \: \: f_s \leqslant f_s^{max}}}}}$

4.) show resultant force exerted by surface on block?

A.

W is the weight of the body,

R is the normal reaction,

$\bold {f_s^{max} }\: is \: the \: limiting \: friction$

P is the applied force and

$\bold{OC} \: is \: the \: resultant \: of \: f_s^{max} \: and \: R$

The angle θ between the normal reaction and the resultant OC is, by definition, the angle of friction.

$tan \theta \: = \frac{bc}{ob} = \frac{oa}{ob} = \frac{ f_s^{max}}{R} \:$

BUT,

$\frac{ f_s^{max}}{R} = \mu_ s \:$

= Coefficient of static friction.

${ \bold{ \boxed {\blue{tan \theta = \mu_s}}}}$

Thus the coefficient of static friction is equal to the tangent of the angle of friction.

5.) relation between angle of friction and coefficient of friction?

A. In the above figure resultant forces

$S = \sqrt{ {F }^{2 + {R}^{2} } }$

$S = \sqrt{ {( \mu \: mg)}^{2} + {(mg)}^{2} }$

$S = mg \sqrt{ { \mu}^{2} + 1}$

When there is no friction (mu=0) S will be minimum

i.e., S = mg

Hence, the range of S can be given by,

$mg \leqslant S \leqslant mg \sqrt{ { \mu}^{2} + 1}$

Answer 2

a person is standing a looking upwards having 45 degree and then hits back 50 and then the angle is 30 degree we have to find the distance of river to the person

◉ GIVEN:-

• angle -45°

• distance taken back- 50m = DC

• shifting back angle - 30°

◉TO FIND:-

• distance of river to the person =CB

◉ SOLUTION:-

$: \longrightarrow \: tan \: 30 \degree \: = \frac{h}{ \:DB }$

$: \longrightarrow \: \frac{1}{ \sqrt{3} } = \frac{h}{ \:DB }$

$\longmapsto h= \frac{DB}{ \sqrt{3} } ....(1.)$

And:-

$: \longrightarrow \:tan45 \degree \: = \frac{h}{CB }$

$: \longrightarrow \:1 = \frac{h}{CB}$

⟼ h = CB ....(2.)

NOW,

(1.) = (2.)

$\leadsto \: \frac{DB}{ \sqrt{3} } = CB$

⇝ DB = ✓3 CB

WE CAN WRITE DB = CB + CD.

⇝ CB +CD= ✓3CB

⇝ CD = ✓3 CB -CB

⇝ CD = CB (✓3-1)

⇝ CD = CB (1.732-1)

⇝ CD = CB(0.732)

NOW WE'LL PUT THE VALUE OF CD = 50m

⇝ 50 = CB (0.732)

$⇝CB = \frac{50}{0.732}$

$\leadsto \:{ \tt{ \pink{ CB = 68.31m}}}$

distance of river to the oerson = CB = 68.31m