while the force of kinetic friction is fk = µ

Transcription

1 19. REASONING AND SOLUION We know that µ s =2.0µ k for a crate in contact with a MAX cement floor. he maximum force of static friction is fs = µ sfn while the force of kinetic friction is fk = µ kfn. As long as the crate is on the cement floor, we can conclude that the magnitude of the maximum static frictional force acting on the crate will alwas be twice the magnitude of the kinetic frictional force on the moving crate, once the crate has begun moving. However, the force of static friction ma not have its maximum value. hus, the magnitude of the static frictional force is not alwas twice the magnitude of the kinetic frictional force.

2 24. REASONING AND SOLUION A circus performer hangs from a stationar rope. Since there is no acceleration, the tension in the rope must be equal in magnitude to the weight of the performer. She then begins to climb upward b pulling herself up, hand-over-hand. Whether the tension in the rope is greater than or equal to the tension when she hangs stationar depends on whether or not she accelerates as she moves upward. When she moves upward at constant velocit, the tension in the rope will be the same. When she accelerates upward, the rope must support the net upward force in addition to her weight; therefore, in this case, the tension in the rope will be greater than when she hangs stationar.

3 26. REASONING AND SOLUION here are three forces that act on the ring as shown in the figure below. he weight of the block, which acts downward, and two forces of tension that act along the rope awa from the ring. Since the ring is at rest, the net force on the ring is zero. he weight of the block is balanced b the vertical components of the tension in the rope. Clearl, the rope can never be made horizontal, for then there would be no vertical components of the tension forces to balance the weight of the block. W

4 34. REASONING In each case the object is in equilibrium. According to Equation 4.9b, ΣF = 0, the net force acting in the (vertical)direction must be zero. he net force is composed of the weight of the object(s)and the normal force exerted on them. SOLUION a. here are three vertical forces acting on the crate: an upward normal force + that the floor exerts, the weight m 1 g of the crate, and the weight m 2 g of the person standing on the crate. Since the weights act downward, the are assigned negative numbers. Setting the sum of these forces equal to zero gives he magnitude of the normal force is F + ( m g) + ( m g) = 0 N 1 2 ΣF = m 1 g + m 2 g = (35 kg + 65 kg)(9.80 m/s 2 )= 980 N b. here are onl two vertical forces acting on the person: an upward normal force + that the crate exerts and the weight m 2 g of the person. Setting the sum of these forces equal to zero gives FN + ( m2g) = 0 ΣF he magnitude of the normal force is = m 2 g = (65 kg)(9.80 m/s 2 )= 640 N

5 37. REASONING AND SOLUION he block will move onl if the applied force is greater than the maximum static frictional force acting on the block. hat is, if F>µ s =µ s mg = (0.650)(45.0 N) = 29.2 N he applied force is given to be F = 36.0 Nwhich is greater than the maximum static frictional force, so the block will move. he block's acceleration is found from Newton's second law. a = ΣF m = F f k m = F µ k mg m = 3.72 m/s 2

6 39. SSM REASONING AND SOLUION Four forces act on the sled. he are the pulling force P, the force of kinetic friction f k, the weight mg of the sled, and the normal force exerted on the sled b the surface on which it slides. he following figures show free-bod diagrams for the sled. In the diagram on the right, the forces have been resolved into their x and components. P sin θ f k θ P f k P cos θ x x mg mg Since the sled is pulled at constant velocit, its acceleration is zero, and Newton's second law in the direction of motion is (with right chosen as the positive direction) F = Pcosθ f = ma = 0 x k x From Equation 4.8, we know that fk = µ kfn, so that the above expression becomes In the vertical direction, Pcosθ µ F = 0 (1) k N F = Psinθ + F mg = ma = 0 (2) N Solving Equation (2) for the normal force, and substituting into Equation (1), we obtain k ( ) Pcosθ µ mg Psinθ = 0 Solving for µ k, the coefficient of kinetic friction, we find P cos θ (80.0 N) cos 30.0 µ k = = = 2 mg Psin θ (20.0 kg) (9.80 m/s ) (80.0N) sin

7 47. REASONING AND SOLUION a. In the horizontal direction the thrust, F, is balanced b the resistive force, f r,ofthewater. hat is, F x =0 or f r =F= N b. In the vertical direction, the weight, mg, is balanced b the buoant force, F b.so gives F =0 F b =mg=( kg)(9.80 m/s 2 )= N

8 50. REASONING he drawing shows the I-beam and the three forces that act on it, its weight W and the tension in each of the cables. Since the I-beam is moving upward at a constant velocit, its acceleration is zero and it is in vertical equilibrium. According to Equation 4.9b, F = 0, the net force in the vertical (or ) direction must be zero. his relation will allow us to find the magnitude of the tension. SOLUION aking up as the + direction, Equation 4.9b becomes 3 + sin sin N = 0 Σ F Solving this equation for the tension gives = 4260 N. W= N

9 54. REASONING he free-bod diagram in the drawing at the right shows the forces that act on the clown (weight = W). In this drawing, note that P denotes the pulling force. Since the rope passes around three pulles, forces of magnitude P are applied both to the clown s hands and his feet. he normal force due to the floor is, and the maximum static frictional force is f MAX s. At the instant just before the clown s feet move, the net vertical and net horizontal forces are zero, according to Newton s second law, since there is no acceleration at this instant. P P W f s MAX SOLUION According to Newton s second law, with upward and to the right chosen as the positive directions, we have + P W = 0 Vertical forces MAX and f s P = 0 Horizontal forces From the horizontal-force equation we find P = f MAX s. But f MAX s = µ s. From the vertical-force equation, the normal force is = W P. With these substitutions, it follows that Solving for P gives MAX P = f s = µ s = µ s ( W P) P = µ s W = ( 0.53) ( 890 N) = 310 N 1+ µ s

10 58. REASONING Since the mountain climber is at rest, she is in equilibrium and the net force acting on her must be zero. hree forces comprise the net force, her weight, and the tension forces from the left and right sides of the rope. We will resolve the forces into components and set the sum of the x components and the sum of the components separatel equal to zero. In so doing we will obtain two equations containing the unknown quantities, the tension L in the left side of the rope and the tension R in the right side. hese two equations will be solved simultaneousl to give values for the two unknowns. SOLUION Using W to denote the weight of the mountain climber and choosing right and upward to be the positive directions, we have the following free-bod diagram for the climber: For the x components of the forces we have + Σ F = sin80.0 sin 65.0 = 0 x R For the components of the forces we have L L 65.0º 80.0 R +x Σ F = cos cos 65.0 W = 0 R L W Solving the first of these equations for R,wefindthat R = L sin 65.0 sin 80.0 Substituting this result into the second equation gives sin 65.0 L cos L cos 65.0 W = 0 or = L W sin 80.0 Using this result in the expression for R reveals that sin 65.0 sin 65.0 = = R L ( ) sin 80.0 W = sin80.0 W Since the weight of the climber is W = 535 N, we find that L R ( ) = W = N = 919 N ( ) = W = N = 845 N