The steepest street in the world is baldwin street in...

The steepest street in the world is Baldwin Street in Dunedin, a small town in New Zealand. It has an angle of 38°. Suppose a trunk is placed on Baldwin Street. For the trunk to remain at rest, a friction force must exist between the pavement of the street and the bottom of the trunk. If an additional force with a magnitude equal to 1/4 of the weight of the trunk is applied uphill in order to hold the trunk in place, what is the coefficient of friction? A: µ=0.464 Now imagine that children are sledding down Baldwin Street after a snowfall. What would be the coefficient of friction if a sled and a rider travel down the hill with a net acceleration of 2 m/s^2? Ans: µ=0.523 The most massive car ever built was the official car of the General Secretary of the Communist Party. The car had a mass of 3335 kg. Suppose this car is moving down a 10° slope when the driver applies the brakes. The wheels stop rotating, and the car slides, accelerating at a rate of -6 m/s^2. Calculate the force of friction acting on the car as it decelerates, and determine from that the coefficient of friction. Ans: µ=0.798 In the sport of grass skiing, the maximum speeds can be quite high. If a skier attained a speed of 26 m/s 6.6 seconds after he started down slope with a 35° incline. Find the coefficient of friction between the grass and the skis. Ans: µ=0.219 In 1989, a hamburger with a mass of 2500 kg was cooked in Wisconsin. If this hamburger slides down a ramp that has a 45° incline, and the coefficient of friction between the hamburger and the ramp is 0.597, what is the net force acting on the hamburger? What is the magnitude of the hamburger’s net acceleration? Ans: Fnet = 6.99 X 10^3; anet = 2.80 m/s^2

Transcript

00:01 Hello everyone, let's proceed to solve this question.

00:02 So this is the scenario given to us in the first part of our question.

00:06 So theta was given us to be 38 degrees.

00:10 All right.

00:11 Now for the truck to be in rest, so for the truck to be in rest, all right? so, the force mg sine of theta should be equal to the applied force, the applied force, that's equal to w by 4 plus the frictional force, plus the frictional force.

00:52 And m g sine theta is the horizontal component of the weight m g so let's see so mg sign of theta will be equal to w and w is the weight of the truck so the weight of the truck is simply mg divided by four so the weight of the truck w is simply m g where m is the mass of the truck g is the excretion due to gravity plus the fractional force that that's the coefficient of friction multiplied by the normal.

01:26 Alright, so from here on simplifying, we get sine theta to be simply equal to 1 by 4 plus mu cos of theta.

01:41 Alright, because n we can substitute to be equal to m g cost of theta.

01:51 So we will substitute this value of n to be here.

01:54 On simplifying we get sine theta to be 1 by 4 plus mu cost theta so finally mu will simply be equal to 4 sine theta minus 1 divided by 4 cos theta all right so let's substitute the values so on substituting the values 4 sine of 38 degrees minus 1 divided by 4 cos of 38 degrees so on simplifying this we get the value of our coefficient of friction to be 0 .464 all right now let's proceed to solve the next part of the question so question number two let's draw the diagram first so theta this is the diagram and theta is given us to be 38 degrees and a net is given us to be 2 meter per second square so from the diagram so we can write the force equation mass multiply by a net that is simply equal to m g sine theta minus the frictional force all right so let's substitute the value in the formulas the formula so m a net will equal to mg sine theta and the frictional force and the frictional force will simply mu times n and from the figure we can see n balances m g phos theta so we can directly write in place of n to be m g force of theta so the formula of frictional force f is simply equal to mu times n all right now for the simplifying we get let's put the values so on simplifying a net is simply equal to g sine theta minus mu g g cos theta so a net was two g was of a so you can take g here 9 .8 so sine of 38 negative of mu of plus 30 all right so i'm simplifying this we take our value of new to be 0 .523 all right so let's move to the third part of the question so let's see the diagram of the third part so the diagram is draw it all right this is the diagram for the third question and we are given the mass of this is the car here so this is the car here and the mass of this car is given us to be 300 three thousand three hundred thirty five kg all right now we're given theta to be ten degrees all right so from the diagram we can write this force equation again so mass multiplied by a net will simply be equal to mg sine of theta and we need to subtract the frictional force from it simply equal to so m multiplied by a net will be equal to mg sine theta and frictional forces again coefficient of friction multiplied by the normal and the normal here is also balancing m g g g g g theta so in place of normal you can directly write mg cost theta all right so value of a net the total acceleration is given as to be negative of six meter per second square so mass gets cancelled so negative of six divided by g that's 9 .8 will simply be equal to sign of 10 degrees negative of mu of 10 degrees all right so on simplifying this we get the value of the coefficient of friction to be 0 .798.

06:27 Alright, so let's proceed to the fourth part of our question.

06:32 So let's draw the diagram of the fourth part first.

06:36 Alright, so this is the diagram for the fourth part.

06:39 Now here, theta is given us, theta is given us to be 35 degrees or bit.

06:46 So it is given that its velocity is 20, 2026 meter per second in time 6 .6 seconds...

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