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General Physics 1, Lec 11 By/T.A. Eleyan 1 Lecture 11 Newton’s Laws of Motion (Applications)

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General Physics 1, Lec 11 By/T.A. Eleyan 2 If the net for force exerted on an object is zero, the acceleration of the object is zero and its velocity remains constant. That is, if the net force acting on the object is zero, the object either remains at rest or continues to move with constant velocity. When the velocity of an object is constant (including when the object is at rest), the object is said to be in equilibrium. Definition of equilibrium The first condition of equilibrium

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General Physics 1, Lec 11 By/T.A. Eleyan 3 Example: A traffic light weighing 122 N hangs from a cable tied to two other cables fastened to a support, as in Figure The upper cables make angles of 37.0 ° and 53.0 ° with the horizontal. Find the tension in the three cables. Solution: Note that the traffic light in equilibrium. Then

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General Physics 1, Lec 11 By/T.A. Eleyan 4 since From the condition of equilibrium By solve the eqs. Example: Suppose the two angles are equal. What would be the relationship between T1 and T2?

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General Physics 1, Lec 11 By/T.A. Eleyan 5 Example: A person weighs a fish of mass m on a spring scale attached to the ceiling of an elevator, as illustrated in Figure. Show that if the elevator accelerates either upward or downward, the spring scale gives a reading that is different from ward, the weight of the fish. Solution:

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General Physics 1, Lec 11 By/T.A. Eleyan 6 We chose upward as the positive y direction (1) If the elevator is either at rest or moving at constant velocity, the fish does not accelerate, and so, Or (Remember that the scalar mg is the weight of the fish.) (2) If the elevator moves with an acceleration a relative to an observer standing outside the elevator in an inertial frame, Newton’s second law applied to the fish gives the net force on the fish: (*)

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General Physics 1, Lec 11 By/T.A. Eleyan 7 For example: if the weight of the fish is 40.0 N and a upward, so that ward, a y = 2.00 m/ s 2, the scale reading from (*) is (2) If a is downward so that a y = - 2.00 m/s 2, then (*) gives us, Thus, we conclude from (*) that the scale reading T is greater than the weight mg if a is upward, so that a y is Positive, and that the reading is less than mg if a is downward, so that, a y is negative.

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General Physics 1, Lec 11 By/T.A. Eleyan 8 Normal Force

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General Physics 1, Lec 11 By/T.A. Eleyan 9 Example: A car of mass m is on an icy driveway inclined at an angle as in Figure. Find the acceleration of the car, assuming that the, driveway is frictionless. Solution:

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General Physics 1, Lec 11 By/T.A. Eleyan 10 Now we apply Newton Newton’s second law in component form, noting that a y = 0: (1) (2) From (1) From (2) Example: solve the example if the mass 1000 kg and the angle 30.

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General Physics 1, Lec 11 By/T.A. Eleyan 11 Example: When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass, as in Figure, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to measure the free-fall acceleration. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord. When Newton’s second law is applied to object m 1, we obtain Solution:

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General Physics 1, Lec 11 By/T.A. Eleyan 12 Similarly, for object m 2 we, find When (2) is added to (1), T cancels and we have Example: find the acceleration and the tension of Atwood's machine in which m 1 =2kg, m 2 =4kg. Then we have

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General Physics 1, Lec 11 By/T.A. Eleyan 13 example: A ball of mass m 1 and a block of mass m 2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass, as in Figure. The block lies on frictionless incline of angle. Find the magnitude of the acceleration of the two objects and the tension in the cord. Equations of motion for m 1 (1) (2) Solution:

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General Physics 1, Lec 11 By/T.A. Eleyan 14 Equations of motion for m 2 (4) (3) Now, solve (2), (3) simultaneously. The acceleration When this is substituted into (2) we find the normal force Example: if m 1 =10kg, m 2 =5kg and =45 find a and T.

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General Physics 1, Lec 11 By/T.A. Eleyan 15 Example: A 3.0 kg mass hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates to the right, the rope makes an angle of 4.0° with the vertical. Find the acceleration of the car. In the vertical direction: Solution:

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General Physics 1, Lec 11 By/T.A. Eleyan 16 In the horizontal direction: Problem: A train is given an acceleration of 5 m/s 2. What is the tension between the two cars with a mass of 2000 kg.

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General Physics 1, Lec 11 By/T.A. Eleyan 17 Forces of Friction In most cases we need to include friction. For many surfaces the force of sliding friction is proportional to the normal force. That is, as the normal force increases, the force of friction increases linearly. When this is true, we write with the proportionality constant called, the coefficient of kinetic friction. This is the friction when the object is moving.

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General Physics 1, Lec 11 By/T.A. Eleyan 18 If I push a table and it doesn't move, there must also be a force pushing back. That force comes from friction, as well. It is the friction when the object is not moving, or static, called the coefficient of static friction. It is always greater than or equal to the coefficient of kinetic friction. It is often harder to start something sliding than to keep it sliding, due to the difference in these coefficients.

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General Physics 1, Lec 11 By/T.A. Eleyan 19 The Static force of friction ( f s ) is the force of friction between two objects when there is no motion. The Kinetic force of friction ( f k ) is the force of friction between two objects when there is motion.

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General Physics 1, Lec 11 By/T.A. Eleyan 20 Example : An 18.0 kg box is released on a 37.0° incline and accelerates down the incline at 0.27 m/s 2. Find the coefficient of friction and the frictional force. (Draw diagram) Forces in y-direction are: Solution :

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General Physics 1, Lec 11 By/T.A. Eleyan 21 Forces in x-direction are: Now, we can relate these to the coefficient of kinetic friction,

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