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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.

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.

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

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

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.

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.

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 ???

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 ?

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.

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.

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

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