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Equilibrium From Princeton Review Book 1. If all of the forces acting on an object balance so that the net force is zero, then: a. the object must be at rest. b. the object’s speed will decrease. c. the object will follow a parabolic trajectory. d. the object’s direction of motion can change, but not its speed. e. none of the above. *E. A is not true UNLESS the object began at rest, but it does not NEED to be true. B cannot be true because a change in speed means acceleration. C cannot be true because objects with no net force follow a straight line at constant speed (Newton’s first law) D cannot be true because a change in direction also requires a force. From Old AP’s Questions 2 and 3 refer to a block of mass m pulled with constant velocity over a floor by a force T inclined at an angle with respect to the floor as shown below. The coefficient of friction between the block and the floor is . 2. The magnitude of the vertical component of the force exerted on the block by the floor is: a. mg. b. mg – Tcos c. mg + Tcos d. mg – Tsin e. Tsin D. The force exerted by the floor (the normal) is equal to the weight MINUS the amount that the person is supporting the block. Proof: F 0 Tsin N - mg 0 (block is not accelerati ng verticall y) (N is normal force) N mg T sin The magnitude of the frictional force is: a. Tcos b. Tsin c. 0 d. mg e. Tcos * A. Since the block is neither speeding up nor slowing down, the two horizontal forces must have the same magnitude, so friction = horizontal component of tension. 4. (1984) A 100 Newton weight is suspended by two cords as shown in the figure below. The tension in the slanted cord is: a. 50 N. b. 100 N. c. 150 N. d. 200 N. e. 250 N. *D. Draw an FBD!!! The weight is in equilibrium in both the horizontal and vertical directions. The weight is purely vertical though, so let’s try that column first: 5. A uniform rope of weight 50 Newtons hangs from a hook as shown above. A box of weight 100 Newtons hangs from the rope. What is the tension in the rope? a. 50 N throughout the rope. b. 75 N throughout the rope. c. 100 N throughout the rope. d. 150 N throughout the rope. e. It varies from 100 N at the bottom of the rope to 150 N at the top. * E. think of the trapeze artist station in the Newton’s third law lab. Every section of rope must support the weight of everything below it. Thus, there is more tension at the top of the rope than at the bottom. This is why most problems involve “massless” ropes. 6. When an object of weight W is suspended from the center of a massless string as shown below, the tension at any point in the string is: a. 2W sin b. W cos 2 c. W cos d. W 2 cos e. W sin *D. Since all of the answers have to do with the weight, analyze the vertical forces: F 0 T cos T cos W 0 Careful - -since the defined angle is relative to the vertical, the vertical component is cosine T W 2 cos 7. A system in equilibrium consists of an object of weight W that hangs from three ropes as shown above. The tensions in the ropes are T1, T2, and T3. Which of the following are correct values of T2 and T3? T2 T3 a. Wtan60 W cos 60 b. Wtan60 W sin 60 c. Wtan60 Wsin60 d. W tan 60 W cos 60 e. W tan 60 W sin 60 * E. Analyze the forces acting on the knot both horizontally and vertically. Also, start with the fact that T1 is equal to the weight. 8. (2004, 51%) An object weighing 300 Newtons is suspended by means of two cords, as shown above. The tension in the horizontal cord is: a. 0 N. b. 150 N. c. 210 N. d. 300 N. e. 400 N. *D 9 (2004B, 75%) A horizontal, uniform board of weight 125 N and length 4 meters is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. What is the tension in the left chain? a. 250 N. b. 375 N. c. 500 N. d. 625 N. e. 875 N. *B Draw a force diagram: