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(II) A large crate of mass 1500 $\mathrm{kg}$ starts sliding from restalong a frictionless ramp, whose length is $\ell$ and whose incli-nation with the horizontal is $\theta$ (a) Determine as a functionof $\theta :(\mathrm{i})$ the acceleration $a$ of the crate as it goes downhill,(ii) the time $t$ to reach the bottom of the incline, (iii) the final velocity $v$ of the crate when it reaches the bottom of the ramp, and (iv) the normal force $F_{\mathrm{N}}$ on the crate. $(b)$ Now assume $\ell=100 \mathrm{m}$ . Use a spreadsheet to calculate andgraph $a, t, v,$ and $F_{\mathrm{N} \text { as functions of } \theta \text { from } \theta=0^{\circ} \text { to } 90^{\circ} \text { in }}$$1^{\circ}$ steps. Are your results consistent with the known resultfor the limiting cases $\theta=0^{\circ}$ and $\theta=90^{\circ} ?$

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a) i. $$a=[g \sin \theta$$ii. $$t=\sqrt{\frac{2 l}{g \sin \theta}}$$iii. $$v=\sqrt{2 g l \sin \theta}$$iv. The normal force $F_{N}$ on the crate is $F_{N}=m g \cos \theta$b) (1500)(9.80) $\cos \theta$, $1.47 \times 10^{4} \mathrm{N}$, $44 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 4

Dynamics: Newton's Laws of Motion

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

University of Winnipeg

McMaster University

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

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So here we're going to draw the incline. We can draw the free body diagram for the crate. This will be our angle, Fada. This would be our crate perpendicular to the surface of contact would be forced, normal, and then going straight down would of course, be force of gravity. So mg Ah, we're going to for part a I we're going to find the acceleration. So sigma f in the ex direction. Let's apply Newton's second law. This would be equal to the X component of the gravity of the gravitational force. This would be equal to M A, and we find that the acceleration is than equal to G sign of data for part two weaken. We want to find the time for a displacement of El so Delta X would be equal to out which would be equal to V X initial t plus 1/2 of a T squared. And so we can eliminate the ex initial. We know that that is going to be zero, so the time would be equal to the square root of two times out over the acceleration or, in this case, G sinus data four. Part three. We're going to say that the final velocity would be equal to the initial velocity plus 80. We know that the initial velocity would be zero. This would be equal to G sign of fada times the square root of two l over G sign of data And so the final velocity would then be equal to the square root of two G. L sign. How's data? At this point, we are going to try to find the ah normal force here. We're going to take the sum of forces in the UAE direction. This is going to be equal to the normal force minus C. Why Components velocity mg co sign of data. This is gonna equal zero because the system has translational equilibrium in the UAE direction. This would be a contusion, anxious. So in the UAE direction, this has ah ah, translational equilibrium. Because of course, the crate is not moving off the surface of the incline. Therefore, forced normal would be equal to m g co sign If Ada now for part B, they're asked. They're telling us that the mass now equals 1500 kilograms. Ah, Al is equaling 100 meters and we know that, uh the acceleration. Now we can plug these in. So the acceleration now becomes 9.809 point 80 Sign of data and the units here would be meters per second squared for tea. This would equal the square root of two times 100 divided by 9.80 Sign of theta the units. Here would be seconds for the force normal, this would be 1500 multiplied by 9.8 uh, co sign of data and the units here would be Newton's. And then for the final velocity, this would be equal to the square root of two times 100 times 9.8 sign if Ada and the units here would be meters per second. And so essentially, we're going to plot all of these, um, as a function of state s. So now plot, uh, acceleration as a function of theta t is a function of data force normal as a function of data and velocity final as a function of data and when we d'oh these air the resulting graphs. So we have acceleration versus angle, the time to the bottom versus angling degrees, the final velocity versus the angle in degrees and then the normal force in Newton's as an angle versus the angle in degrees. Now you want to check the limiting cases at zero degrees. The acceleration and the final velocity should be zero, which checks out at 90 degrees. Uh, the normal. Ah, and at zero degrees. Rather, the normal force should be equal to the weight, and it's going to be a little bit less than 15,000 Newtons. So this checks out as well. And then the time to the bottom uh, zero degrees should be approaching infinity because it should, of course, already be had the bottom. So this checks out because here we have a vertical Assen toque at, uh zero at zero degrees. So here the zero degree limiting case works out than 90 degrees. The 90 degree limiting case works as well. The acceleration should be equal to G, the acceleration due to gravity. And as you can see here, it's a little bit less than 10. So this checks out here. The final velocity should be between 40 but around 45 meters per second, so that checks out. The normal force should be zero because there shouldn't be any contact with any other surface or 90 degrees and normal force should be zero Newton's. This works out and then the time to the bottom should go should decrease as Thea and GLE increases. This works out, and at 90 degrees the time to the bottom will be approximately four and 1/2 seconds. That is the end of the solution. Thank you for watching.

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