static friction

Q. The coefficient of static friction between the tires of a car and a dry road is 0.62. The mass of the car is 1500 kg. What maximum braking force is obtainable on an 8.6degree downgrade ?

B = ?
Friction = 1500(9.81)cos(8.6) * (0.62) = 9020.72 N
and the component of weight is also acting down the slope = 1500(9.81)sin(8.6)=2200.41N

now I don't understand how to take the directions ..the friction would act opposite to the braking force but what about the 2200.41 N force ..would it also act opposite to the braking force or will we consider the 2200.41 N force in the first place because the answer in the bookis suppose to be 9.0kN.

Reply 1

Drummy

10

there are 3 forces acting on the car

its weight vertically downwards
the reaction force acting out of the slope
friction acting up the slope

to generate the equation we need resolve perpendicular to the slope which gives

R = mg cos\theta

so

F = \mu R = \mu mg cos\theta

bung in the numbers and Robert's yer father's brother

Reply 2

Theorist

10

e-n-i-g-m-a

now I don't understand how to take the directions ..the friction would act opposite to the braking force but what about the 2200.41 N force ..would it also act opposite to the braking force or will we consider the 2200.41 N force in the first place because the answer in the bookis suppose to be 9.0kN.

If I'm right, the braking force is mainly because of friction.

Therefore, the braking force is simply the difference between the friction acting on the car and the component of its weight parallel to the plane.

Reply 3

Drummy

10

The braking force IS the friction acting up the plane

if you use my equation you get the textbook answer

the car is static, i.e. in equilibrium. The braking force (the friction) is stopping it from sliding down the slope

Reply 4

Theorist

10

Drummy

The braking force IS the friction acting up the plane

if you use my equation you get the textbook answer

the car is static, i.e. in equilibrium. The braking force (the friction) is stopping it from sliding down the slope

Again, my bad. :smile:

Reply 5

e-n-i-g-m-a

OP

why are we ignoring the component of weight that acts down the slope. Even when the mass is stationary the component of weight is acting ???

Reply 6

Drummy

10

you can resolve in any directions, not just vertically and horizontally

I resolved perpendicular to the slope. The component of the weight down the slope therefore does not appear in the equation.

If I resolved up/down the slope then this would give

F = mg sin\theta

Reply 7

e-n-i-g-m-a

OP

so if ur considering down and up the slope...You have friction acting up the slope and the component of weight acting down the slope...what happens to teh braking force then ?

Reply 8

Drummy

10

your definitely reading too much into this question m8

There are just 3 forces acting on the object and as it is static they are in equilibrium.

W, The weight of the object downwards

R, The reaction force OUT of the slope (normal contact force)

F, Friction UP the slope

Whatever direction we choose to resolve in there is no resultant force. By resolving parallel and perpendicular to the slope we get equations which either dont have R or dont have F. Having 2 unknowns would make it harder to solve.

Reply 9

e-n-i-g-m-a

OP

really sorry to bother you again ....resolving parallel to the plane ..we do get the equation mgsin theta

static friction

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