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(III) A curve of radius 68 $\mathrm{m}$ is banked for a design speed of85 $\mathrm{km} / \mathrm{h} .$ If the coefficient of static friction is 0.30 (wet pave-ment), at what range of speeds can a car safely make thecurve? [Hint: Consider the direction of the friction forcewhen the car goes too slow or too fast.]

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$$17 \mathrm{m} / \mathrm{s} \leq v \leq 32 \mathrm{m} / \mathrm{s}$$$$61 \mathrm{km} / \mathrm{h} \leq v \leq 115 \mathrm{km} / \mathrm{h}$$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

University of Sheffield

University of Winnipeg

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hi radios off the bank roll here is given us R is equal to 68 meter. This road is bent for a design speed off 85 kilometer per hour means the speed which is possible without considering any friction over the bank surface. And we know for this speed the angle off the banks surface is given by the formula. Tan theta is equal to the square by our jeep in this formula. In this expression, we use no friction off the bench surface, so plugging all the known values here. But before that we will convert this kilometer per are into meter per second, for which we will multiply it by five by 18 and it will be converted into meter per second, which comes out Toby, 23.6 meter or second. So putting this value here this is a square off 23.6, divided by radius, which is 68 m into G, which is 9.8 and here it approximately comes out to be 0.836 means the tangent of the angle off this bank wrote, which it makes with the horizontal comes out to be 0.836 and it will be used as such in finding the range off the velocity is the optimum speech off the car off the vehicle at which it can cover this bankroll. So four, the maximum speed at which the car safely make the girl or the maximum speed off the car. This car will be having a tendency to slide up the inclined here. This is the bankroll, then for the maximum speed. If you consider the maximum speed of the car, this car will be having a tendency to slide up as well. It will be having attend institute tendency to slide up so force of friction between its tire and the road will be acting downward. In that case, if we find ReMax, it has an expression. It is given by an expression R G bracket mu plus Dan Tita, divided by one minus mu Dan Tita. Similarly, for the minimum speech off the car, what a minimum speed on the car will be moving at the minimum speech over this curved path. It will be having a tendency to slide down and when the car will be having a tendency to slide down Now the force of friction will act in upward direction and in that case, minimum speech required will be given by the expression square root are G Dante Dah minus mu divided by one plus mu then Tito. So now plugging in all the known values in these expressions. For first of all, the max here this we Max will be given by for our this is 68 m for G. This is 9.8 m for second square for new this waas 0.3 plus 10 theater. We know we have found its value already This is 0.836 So here this is 0.836 divided by one minus mute Antietam means one minus 0.3 into again 0.836 So it becomes 68 multiplied by 9.8 into one point 136 in numerator adding this 0.3 in 0.836 and divided by one minus this, it comes out to be 0.7492 So finally it becomes 101,000 and 10 0.45 And that's why I wrote off this so Finally, we Max here comes out to be 31.8 meter per second or approximately. We can say this is 32 meter per second, the maximum optimist optimum speech with which the car can take this circular turn now for the minimum again, putting all these known values for radius. This is 68 m for G. This is 9.8 m per second squared, then 0.836 for 10 theta minus new 0.3 divided by one plus new means 0.3 multiplied by 10 theta means 0.836 and then solving them. This is 68 in 29.8 into 0.536 divided by 1.2508 which comes out to be 285.6. So finally, this is 16.9 meter per second or approximately. This is 17 m per second hands. The optimum speed off the car in order to take a safe turn over this bank wrote, is lying between 17 m per second. Do 32 m per second and it becomes the answer for this given problem. Thank you

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