1. The coefficient of static friction at C is μs = 0.3. Neglect the thickness of the ring. (Figure 1) Determine the largest magnitude of the force F that can be applied to the 4.4-kg ring, without causing it to slip.
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1. The coefficient of static friction at C is μs = 0.3. Neglect the thickness of the ring. (Figure 1) Determine the largest magnitude of the force F that can be applied to the 4.4-kg ring, without causing it to slip.
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- Question 1 25 A 380 kN horizontal force P is applied to a 210 kN box resting on an 25° incline. The line of action of Ppasses through the center of gravity of the box. The box Is 3.5 m wide x 3 m tall, and the coefficents of friction between the box and the surface are 14 -0.6 and 14% -0.51. Determine the magnitude of the friction force acting on the box. F'= The box is In static equilibrium Sliding up the incline Sliding down the incline. ○ TippingStatics 1 IYGBX A helium balloon floats at the end of a string. The net lift force (the buoyancy force minus the weight) on the balloon is 33_N. If the tension in the string exceeds 50 N of force, the string will break. How great a force can a horizontal wind exert on the balloon before the string breaks? Flift Fwind 24. Fwind ABC 1--1 44.69 N 39.32_N 41.29_N Tstring D. 46.76_N 42.63_N 37.56 N DEE E. F.34.
- Incorrect Calculate the horizontal force P on the light 10° wedge necessary to initiate movement of the 34-kg cylinder. The coefficient of static friction for both pairs of contacting surfaces is 0.26. Also determine the friction force Fg at point B. (Caution: Check carefully your assumption of where slipping occurs.) Answers: P= i7.022 Fa= 22.155 34 kg B ! N ! N5. In a carnival ride called the Giant Swing, a seat is connected to two cables, one of which is horizontal (see figure). The seat swings in a horizontal circle with a frequency of 18.0 rev/min. If the seat weighs 205 N and an 655-N person is sitting in it, find the tension in each cable. 40.0⁰ -7.50 m- >5) A box with mass m=1.25 kg rests on the top of a bar. The coefficient of static friction between the box and the bar is μs=0.73 and the coefficient of kinetic friction is μk=0.39. a) Write an expression for Fm the minimum force required to produce movement of the box on the top of the bar. b) Solve numerically for the magnitude of the force Fm in Newtons. c) Write an expression for a, the box's acceleration, after it begins moving. (Assume the minimum force, Fm, continues to be applied.) d) Solve numerically for the acceleration, a in m/s².
- 3. Why the coefficient of kinetic friction is dependent on property of the surface not the weight of block?1) a 15kg box is resting on a horizontal surface a) what is the minimum horizontal force required to make the box to begin to slide. If the coefficient of static friction is 0.35 b) What is the acceleration of the system if a person pushes the box with a force of 90N. If the coefficient of the kinetic friction is 0.20.( Take g as 9.8m/s)3. A rope ABCD is looped over two pipes as shown. Knowing that the coefficient of static friction is 0.25, determine (a) the smallest value of the mass m for which equilibrium is possible; (b) the largest value of mass m for which equilibrium is possible. A B 50 kg 30% m C D
- 9. Two blocks are positioned on surfaces, each inclined at the same angle of 58.3 degrees with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines as shown, so the blocks can slide together. The mass of the black block is 3.66 kg, and the coefficient of kinetic friction for both blocks and inclines is 0.460. Assume static friction has been overcome and that everything can slide. What is must be the mass of the white block if both blocks are to slide to the LEFT at an acceleration of 1.5 m/s^2? 2.16 kg 9.99 kg 6.02 kg 3.06 kgQuestion 15 The force of sliding friction depends on the coefficient of friction and; A the weight of the object. B the area of the surface in contact. the mass of the object. D the force perpendicular to the surface.A2 Kg object moves on a track to the right at 10 m/s, and then reaches a circular depression. Assume the object stays on the track, and the radius of the track is 4 meters at the bottom of the circular depression. Also assume the bottom is 4 meters below the initial position of the object. There is no friction. 2 Kg 10 m/s When the object reaches the bottom of the circular depression, calculate the normal force that the track exerts on the object.