A truck is traveling down a road with a 4 percent grade at a...

A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.5 mi/h. An antiskid braking system limits the braking force to a value at which the wheels of the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels is 0.60, determine the shortest time needed for the truck to slow down.

Transcript

00:01 In this question, it is given that a truck is traveling down a row with a 4 % rate.

00:11 So we can calculate the angle theta.

00:16 So this is the angle theta and theta can be written as tan inverse 4 by 100.

00:25 And tan inverse 4 by 100 can be calculated as 2 .29 degree.

00:31 This is the angle theta and consider this is the truck.

00:40 So the truck is moving on an inclined plane at an angle theta and consider m be the mass of truck and the truck is moving with a speed that is equal to 75 miles per hour and the truck the truck it brakes are applied to slow it down to a speed that is the final speed equal to 22 .5 miles per hour and it is given that an anti -skid braking system limits the braking force to a value at which the wheels of the truck are just above to slide and it is given that the kinetic friction coefficient of kinetic friction between the road and the wheels of the truck is 0 .60 and we need to determine the shortest time needed for the truck to slow down so we need to find the interval of time.

02:01 So i am drawing the free -body diagram here the truck is moving with the speed initial speed that is v -i so it will be in this direction mvi and the normal force acting on the truck is like this in this direction and this will be the weight acting on the weight of the truck mg and these are the horizontal and vertical components so this will be angle theta and the vertical component that will be mg cos theta and the horizontal component will be mg sine theta.

03:00 So there will be a frictional force acting along this direction opposite to the displacement.

03:11 And we know that the friction of force fk that will be equal to mu k times n...

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