Answer:
the direction of the friction force on the car is in the same direction to the VELOCITY of the car.
Explanation:
When the brakes are applied on a moving car then the speed of the car reduces slowly. The slowing of speed and the stopping of car takes a certain time which is proportional to the force applied by the brakes.
The brakes when applied to the moving part on the axle of the vehicle be it either drum or the disc offers the resistance to the rotation of the wheels in the form of friction.
As we know that friction always acts in a direction opposite to the relative motion between the two surfaces. So does here in this case the friction of brakes acts in a direction tangentially opposite to the rotation at the point of application of the brakes.
Similarly, the friction between the tyres and the road acts in a direction tangentially opposite to the direction of the relative motion between the surfaces i.e. the tyres move backwards relative to the road surface and hence the friction acts in the forward direction and is in the same direction as the motion of the car.
Answer:
a) Fi = 85.76 N
b) Fi = 87.8 N
Explanation:
Given:-
- Density of hydraulic oil, ρ = 804 kg/m^3
- The radius of input piston, ri = 0.00861 m
- The radius of output piston, ro = 0.141 m
Find:-
What input force F is needed to support the 23000-N combined weight of a car and the output plunger, when
(a) the bottom surfaces of the piston and plunger are at the same level
(b) the bottom surface of the output plunger is 1.10 m above that of the input plunger?
Solution:-
For part a.
- We see that both plungers are equal levels or there is no pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers to be equal. This part is an application of Pascal's Law:
Pi = Po
Fi / Ai = Fo / Ao
Fi = Ai / Ao * Fo
Fi = (ri/ro)^2 * Fo
Fi = ( 0.00861 / 0.141 )^2 * 23000
Fi = 85.76 N
For part b.
- We see that both plungers are at different levels so there is pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers plus the differential in heads. This part is an application of Bernoulli's Equation:
Pi = Po + ρ*g*h
Where, h = Elevation head = 1.10m
Fi = (ri/ro)^2 * Fo + ρ*g*h*π*ri^2
Fi = 85.76 + (804*9.81*1.1*3.142*0.00861^2)
Fi = 87.8 N
Answer: 0.9
Explanation:
For an inclined surface the coefficient of friction (n) is the ratio of the moving force (Fm) to the normal reaction (R) acting on the body.
n = Fm/R
Fm = WSintheta = mgsintheta
R = Wcostheta = mgcostheta
n = Wsintheta/Wcostheta
n = sintheta/costheta
n = tan theta
n = tan 42°
n = 0.9
Therefore, minimum coefficient of static friction between the vehicle’s tires and the road is 0.9
Answer:
5.4 m
Explanation:
mass, m = 34 kg
initial velocity, u = 9 m/s
final velocity, v = 0 m/s
coefficient of friction, μ = 0.6
Angle of inclination, θ = 10°
Let teh distance traveled before to come into rest is d.
According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the block.
Work done by the gravitational force = W1 = - mg Sinθ x d
Work done by the frictional force = W2 = - μ N = - μ mg Cosθ x d
negative sign shows that the direction of force and the direction of displacement is opposite to each other.
Total work done W = W1 + w2
W = - 34 x 9.8 x Sin 10 x d - 0.6 x 34 x 9.8 x cos 10 x d
W = - 254.86 d
Change in kinetic energy = 0.5 x m (v^2 - u^2)
= 0.5 x 34 (0 - 81) = - 1377
So, W = change in KE
- 254.86 d = - 1377
d = 5.4 m
The law of conservation of mass<span> states that </span>mass<span> in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the </span>law of conservation of mass<span>, the </span>mass<span> of the products in a chemical reaction must equal the </span>mass<span> of the reactants.
</span>
Every chemical equation<span> adheres to the </span>law of conservation of mass<span>, which states that </span>matter<span> cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical </span>equation.