Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details
Get extra R 700 off



I have a query?

Can you tell me which is the direction of static friction acting on us when we do run over a treadmill when

a) we are at rest with respect to ground

b) we are accelerating with respect to ground either in forward or in backward direction of our run.

6 years ago


Answers : (1)


Dear siddharth,


Static friction is the friction experienced when we try to move a stationary body on a surface, without actually causing any relative motion between the body and the surface which it is on.

It can be defined as the force of friction which exactly balances the applied force during the stationary state of the body.

Static frictional force is a self-adjusting force. i.e., the static friction will always be equal and opposite to the force applied.


static friction

Consider the figure, R is the reaction force due to the weight W. fr is the friction and F the external force. F = -fr when no motion occurs.

Laws of static friction

  • The maximum force of static friction is independent of the area of contact.

  • The maximum force of static friction is proportional to the normal force i.e., if the normal force increases, the maximum external force that the body can withstand without moving, also increases.

We are all IITians and here to help you in your IIT JEE preparation.

All the best.

 If you like this answer please approve it....

win exciting gifts by answering the questions on Discussion Forum


Sagar Singh

B.Tech IIT Delhi

6 years ago

Post Your Answer

Other Related Questions on Mechanics

A particle is moving in straight line such that its velocity changes with time t as v=t^2-t. The time interval for which particle retard is
Differentiate v w.r.t t time. dv/dt = 2t – 1 = a for retardation a should be negative. that is to say that, (2t – 1) should be less than zero or t should be less than 0.5 s. thus time...
Vikas TU 2 months ago
A body is dropped from a height of 16 m The body strikes the ground and looses 25% of its velocity. The body rebounds to a height of ........
Let v be the velocity when it stri​kes the ground .ie final velocity ,then ​ v ​2 -0 ​2 ​=2*16*10=320,after rebound velocity V=3/4v,hight attained h=V ​2 ​/2*10=(3/4) ​2 ​...
abhi 3 months ago
When particle is dropped it will reached at ground with velocity √2gh OK.then it will rebounded with same velocity but of lose of energy it`s velocity becomes 0.75√2gh then after rebounded...
Praveen Mishra 2 months ago
2kg of gas at a pressure of 1.5 bar.occupies a volume of 2.5m cube.if this gas compresses isothermally to 1/3 times the initial volume.find temperature,work done,heat transfer.
As the pocess is isothermal chanege in Internal energy would be zero and hence temperature remains costant that is T. From ideal gas eqn. PV =nRT 1.5 * 2.5*1000 = 2000/32 * 0.0821 * T we...
2017 years ago
If T 1 is the time of up the ball to maximum height and T 2 is the of down the ball from maximum height to ground. Then which is one greater If “Air Drag” is also considered or both (T 1...
Dear Vikas, U should also consider ‘Air Drag’. the way us solve the solution is seems like the general sase of free falling. and in free falling case both T 1 and T 2 Are same. And there is ...
Kuldeep Pal one month ago
for upward, h = uT1 – 0.5gT1^2................(1) for downward, h = 0 + 0.5gT2^2..........(2) comparing both the eqns. it seems, h is more for 2 nd eqn. therfore T2 would be more. for...
Vikas TU one month ago
Two persons holding rope of negligible wt. at its horizontal ends . A 15 kg weight is attached to mid of rape which no longer remain horizontal. Minium tension required to straighten rope...
At the mid of the rope the weight will be makingof 45 degree agle with x and y axis both. Therefore components the Tension force as Tcos45 and Tsin45 in x and y direcn. respectively. thus...
Vikas TU 2 months ago
Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 11.4m.The diameter of the moon is 3450km and the distance between the earth and the moon is...
f=11.4,u=3.8*10^5 km so u>>f ....then v=f=11.4. m=-v/u =11.4m/3.8*10^8 m =-3*10^-8. now| m|=hi/ho ,,,3*10^-8 =hi/3.45*10^6,,,,,,hi=10.35*10^-2,,,,,hi=10.35mm I think,it clear your doubt
View all Questions »

  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
Get extra R 650 off

  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details

Get extra R 700 off

More Questions On Mechanics

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!!
Click Here for details