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Exercise Class For College Physics 俞颉翔(Jiexiang Yu) 2010-10-20 Email: [email protected] Office: 2401, East Guanghua Building Problem 4.58 on P113 Consider the 52kg mountain climber. a) b) Find the tension in the rope (T) and the force that the climber must exert with her feet on the vertical rock face to remain stationary (Fl). What is the minimum coefficient of friction between her shoes and the cliff? For x direction: T sin 31o Fl cos15o So sin31 o Fl T 0.533T o cos15 For y direction: 31o T cos 31o Fl sin 15o mg T 0 sin 31 0 T (cos 31 sin 15 ) mg 0 cos 15 Then we have : y Fl 0 T 1.005mg 512.1N Fl 0.533T 273.0 N x 15o W=mg Problem 4.58 on P113 ——Sine Law a b c 2R sin A sin B sin C R refers to the radius of circumscribed circle in a triangle. Another solution Since the system remains stationary, the additions of the force vectors is zero. Fl mg T sin 74o sin 75o sin 31o sin 75o T mg 512.1N o sin 74 sin 31o Fl mg 273.0 N o sin 74 31o T W=mg 75o 15o 74o Fl 74o 75o 31o 180o Problem 4.58 on P113 Coefficient of friction N Fl cos 15o f Fl sin 15o Then the s is given by f s tan 15o 0.268 N f 15o Fl N Something about friction Coefficient of static friction, μs Coefficient of kinetic friction,μk. Usually, μs > μk in the same situation Problem 1 A horizontal force F = 12N pushes a block weighing 5N against a vertical wall. μs = 0.60 and μk = 0.40. Assume the block is not moving initially. Will the block start moving? F W N = F =12N The maximum of static friction is: f s max s N 7.2N W So the block remains stationary. f N W W F Problem 2 F Someone exerts a force F directly up on the axle of the pulley. Consider the pulley and string to be mass-less and the bearing frictionless. Two object, m1=1.2kg and m2=1.9kg, are attached to the opposite ends of the string, which passes over the pulley. The m2 is in contact with the floor. a) b) c) Find the largest value of F may have so that m2 will remain at rest on the floor. what is the tension in the string if the upward force F is 110N? With the tension determined in b), what is the acceleration of m1? m1 m2 a) Since the pulley is mass-less, the net force exerted on it is zero. Thus the tension in the string: 1 T F 2 Then we get the maximum of the tension: F T T Tmax m2 g and the maximum of F: Fmax 2Tmax 2m2 g 2 1.9 9.8 37.24N T m2 W2 = m2g b) The tension in the string: 1 T F 55 N 2 c) T m2 W2=m2g The acceleration of m1: T m1 g 55 a1 9.8 36.03m/s 2 m1 1.2 T m1 W 1= m1g Problem 3 The two blocks, m = 16kg and M = 88kg are free to move. The coefficient of static friction between the blocks is μs = 0.38, but the surface beneath M is frictionless. Find the minimum horizontal force F required to hold m against M. F m M No friction Acceleration of the two blocks: F m F a mM No friction Acceleration of the m block: FN a m M N f F m W=mg The friction : f mg s N N’ Acceleration of the M block: N a M M No friction f N W=Mg From the same acceleration of the two blocks, we have: F m No friction M N F mM Put this into the function of the friction f: f mg s N s M N f F m W=mg M F mM So N’ mM 16 88 F mg 16 9.8 487.66 N M s 0.38 88 No friction M f N W=Mg Problem 4 You throw a ball with a speed of 25.3m/s at an angle of 42.0o above the horizontal directly toward a wall. The wall is 21.8m from the release point of the ball. a) b) c) d) how long is the ball in the air before it hits the wall? how far above the release point does the ball hit the wall? What are the horizontal and vertical components of its velocity as it hits the wall? Has it passed the highest point on its trajectory when it hits? 25.3m/s 42.0o 21.8m Solution a) The time taken for the ball to hit the wall: l 21.8 t 1.16s o v0 cos 25.3 cos 42 b) The Vertical distance above the release point as the ball hits the wall: 1 2 y v0 sin t gt 13.04m 2 25.3m/s 42.0o 21.8m Solution c) The vector of velocity as it hits the wall: vx vx 0 v0 cos 18,8m/s v y v y 0 gt v0 sin gt 5.57m/s d) Since vy > 0, the ball hasn’t passed the peak point of its trajectory. 25.3m/s 42.0o 21.8m Problem 5 A chain consisting of five links, each with mass 100g, is lifted vertically with a constant acceleration of 2.5m/s2. Find: a) b) c) The forces acting between adjacent links, The force F exerted on the top link by the agent lifting the chain, The net force on each link. F 5 4 3 2 1 For link 1: a t 21 w , t 21 m(a g ) 0.1 (2.5 9.8) 1.23N m For link 2: a t32 t12 w , t32 m(a g ) t 12 0.1 (2.5 9.8) 2 2.46 N m For link 3: For link 4: a t54 t34 w , t54 m(a g ) t 34 0.1 (2.5 9.8) 4 4.92 N m For link 5: t t w a 43 23 , t 43 m(a g ) t 23 0.1 (2.5 9.8) 3 3.69 N m F t 45 w a , F m(a g ) t 45 0.1 (2.5 9.8) 55 6.15N m F 5 4 3 2 1 The net force on each link: Fn ma 0.1 2.5 0.25N F Question: A vertical force F is exerted on a rope of which mass is M and length L with a constant acceleration of a. Find the tension as a function of the distance x from the bottom of the rope. x