ism_chapter_05

# ism_chapter_05 - Chapter 5 Applications of Newtons Laws...

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265 Chapter 5 Applications of Newton’s Laws Conceptual Problems 1 Determine the Concept Because the objects are speeding up (accelerating), there must be a net force acting on them. The forces acting on an object are the normal force exerted by the floor of the truck, the weight of the object, and the friction force; also exerted by the floor of the truck. Of these forces, the only one that acts in the direction of the acceleration (chosen to be to the right in the free-body diagram) is the friction force. . accelerate object to the causes that force the be must truck the of floor the and object e between th friction of force The *2 Determine the Concept The forces acting on an object are the normal force exerted by the floor of the truck, the weight of the object, and the friction force; also exerted by the floor of the truck. Of these forces, the only one that acts in the direction of the acceleration (chosen to be to the right in the free-body diagram) is the friction force. Apply Newton’s 2 nd law to the object to determine how the critical acceleration depends on its weight. Taking the positive x direction to be to the right, apply Σ F x = ma x and solve for a x : f = µ s w = s mg = ma x and a x = s g same. the are ons accelerati critical the and of t independen is Because w, m a x

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Chapter 5 266 3 Determine the Concept The forces acting on the block are the normal force n F r exerted by the incline, the weight of the block g r m exerted by the earth, and the static friction force s f r exerted by an external agent. We can use the definition of µ s and the conditions for equilibrium to determine the relationship between s and θ . Apply x x ma F = to the block: f s mg sin = 0 (1) Apply y y ma F = in the y direction: F n mg cos = 0 (2) Divide equation (1) by equation (2) to obtain: n s tan F f = Substitute for f s ( s F n ): s n n s tan = F F and correct. is ) ( d *4 Determine the Concept The block is in equilibrium under the influence of , n F r , m g r and ; s f r i.e., n F r + g r m + s f r = 0 We can apply Newton’s 2 nd law in the x direction to determine the relationship between f s and mg . Apply 0 = x F to the block: f s mg sin = 0 Solve for f s : f s = mg sin and correct. is ) ( d
Applications of Newton’s Laws 267 5 •• Picture the Problem The forces acting on the car as it rounds a curve of radius R at maximum speed are shown on the free-body diagram to the right. The centripetal force is the static friction force exerted by the roadway on the tires. We can apply Newton’s 2 nd law to the car to derive an expression for its maximum speed and then compare the speeds under the two friction conditions described. Apply = a F r r m to the car: = = R v m f F x 2 max max s, and = = 0 n mg F F y From the y equation we have: F n = mg Express f s,max in terms of F n in the x equation and solve for v max : gR v s max µ = or s max constant = v Express ' v max for s 2 1 s = ' : max max s max % 71 707 .

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ism_chapter_05 - Chapter 5 Applications of Newtons Laws...

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