The man tries to reposition the 50 kg refrigerator in his kitchen by pushing it from rest with a force P = 170 N, inclined 30° from the horizontal at 1.65 m from the floor. The coefficient of non-sliding friction between the floor and the refrigerator is μs = 0.40.
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Need answers asap.
The man tries to reposition the 50 kg refrigerator in his kitchen by pushing it from rest with a force P = 170 N, inclined 30° from the horizontal at 1.65 m from the floor. The coefficient of non-sliding friction between the floor and the refrigerator is μs = 0.40.
After pushing, the refrigerator started to tip.
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- A 3.2 kg flagpole extends from a wall at an angle of 25° from the horizontal. Its center of gravity is 1.6 m from the point where the pole is attached to the wall. What is the gravitational torque on the flagpole about the point of attachment?When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure a. The total gravitational force on the body, Fg, is supported by the force n exerted by the floor on the toes of one foot. A mechanical model of the situation is shown in Figure b, where T is the force exerted by the Achilles tendon on the foot and R is the force exerted by the tibia on the foot. Find the values of T, R, and ? when Fg = n = 720 N. (For ?, enter the smaller of the two possible values between 0° and 90°.) T should be in the 1700-1800 range, please be careful and not give a wrong answer, thanks :)d Ꮎ 2. A uniform ladder of mass 25.0 kg and length 10.0 m leans against a frictionless wall at an angle = 65.00 with respect to the ground as shown. The coefficient of static friction between the ladder and the ground is 0.350. What maximum distance (d) can an 75.0 kg worker climb the ladder without it slipping? O 5.00 m O 9.97 m O 10.0 m 8.34 m
- John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high (see the figure below). The handles make an angle of 0 = 20.0° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 407 N is exerted at the center of the wheel, which has a radius of 18.0 cm. Assume the brick remains fixed and does not slide along the ground. Also assume the force applied by John is directed exactly toward the center of the wheel. (Choose the positive x-axis to be pointing to the right.) & (a) What force (in N) must John apply along the handles to just start the wheel over the brick? 1690.53 X Your response differs from the correct answer by more than 10%. Double check your calculations. N (b) What is the force (magnitude in kN and direction in degrees clockwise from the -x-axis) that the brick exerts on the wheel just as the wheel begins to lift over the brick? magnitude KN direction ° clockwise from the -x-axisA 200 N sign hangs from the middle of a cable between two buildings. The ends of each cable are attached to each building at the same height. The cable cannot exceed 1500 N without breaking. Find the minimum angle that the cable can make from the horizontal.A horizontal force of 170 N is required to push a 180-kg refrigerator across a level linoleum floor. The refrigerator rolls on four 50-mm-diameter plastic wheels. Neglect all friction except rolling resistance and determine the coefficient of rolling friction between the plastic wheels and the linoleum floor.
- A brick of mass 1.19 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle has a radius of 1.34 m. The cable will break if the tension exceeds 45.0 Newtons. What is the maximum speed the brick can have at the bottom of the circle before the cable will break?A mixing beater consists of three thin rods, each 10.6 cm long. The rods diverge from a central hub, separated from each other by 120°, and all turn in the same plane. A ball is attached to the end of each rod. Each ball has cross-sectional area 4.20 cm? and is so shaped that it has a drag coefficient of 0.620. The drag force on each ball is R = Dp A v2 where D is the drag coefficient, p the density of the fluid, A the cross-sectional area, and v the speed of the object moving through the fluid. (a) Calculate the power input required to spin the beater at 1000 rev/min in water. w (b) The beater is taken out of the water and held in air. If the input power remains the same (it wouldn't, but if it did), what would be the new rotation speed? rev/minChapter 12, Problem 037 GO In the figure, a uniform plank, with a length L of 6.83 m and a weight of 386 N, rests on the ground and against a frictionless roller at the top of a wall of height h = 2.78 m. The plank remains in equilibrium for any value of e = 70.0° or more, but slips if e < 70.0°. Find the coefficient of static friction between the plank and the ground. Roller Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work
- When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure P8.24a. The total gravitational force on the body, F, is supported by the force n exerted by the floor on the toes of one foot. A mechanical model of the situation is shown in Figure P8.24b, where T is the force exerted by the Achilles tendon on the foot and R is the force exerted by the tibia on the foot. Find the values of T, R, and 0 when F, = n = 700. N. -Achilles tendon Tibia 15.0° 18.0 cm 25.0 cm Figure P8.24Question 8A 45.5‑kg box with the dimensions a×b (the width a = 0.6 m and height b = 1.3 m) is standing on a rough horizontal floor (μs = 1.3). A Professor wants to tip the box over, she pushes on the right side of the box with a horizontal force F = 140 N at a distance h = 1.04 m above the floor as shown in the picture below. The other forces acting on the box are the gravity force (applied to the CoM of the box), the static friction force Fs and the normal force N (both applied to the lower right corner of the box, point Q). Calculate the magnitude and direction of the torques produced by the forces acting on the box relative to the lower right corner (point Q). Assume uniform mass distribution. 1. The torque due to the external force, τQ(due to F) = 2. The torque due to the weight of the box, τQ(due to mg) = 3. The torque due to the friction force, τQ(due to Fs) = 4. The torque due to the normal force, τQ(due to N) =