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Physics for Scientists and Engineers: Foundations and Connections
- Problem 3: A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Consider the Boeing 787 "Dreamliner", with a mass of 189000 kg. In this particular model, the distance from the front wheels to the rear set of wheels is 21 m. Part (a) If the center of mass of the airplane is along a line through the center and 3.00 m in front of the rear wheels, how much force, in meganewtons, does the ground exert on each set of rear wheels when the plane is at rest on the runway? F, = sin() cos() tan() 8 9 HOME cotan() asin() acos() E 5 6 atan() acotan() sinh() 2 3 cosh() tanh() cotanh() END ODegrees O Radians Vol BACKSPACE DEL CLEAR Submit Feedback I give up! Hint Part (b) How much force, in meganewtons, does the ground exert on the front set of wheels?arrow_forwardA ladder leans against a wall 5 meters above the ground so a homeowner can clear his gutters of leaves from last fall. The bottom of the ladder just begins to slip when it is 1.7 meters from the wall. What is the coefficient of static friction between the bottom of the ladder and the ground?arrow_forwardA 1 470-kg automobile has a wheel base (the distance between the axles) of 3.10 m. The automobile's center of mass is on the centerline at a point 1.00 m behind the front axle. Find the force exerted by the ground on each wheel. each front wheel: each rear wheel:arrow_forward
- A pencil rests against a corner, as shown below. The sharpened end of the pencil touches a smooth vertical surface and the eraser end touches a rough horizontal floor. The coefficient of static friction between the eraser and the floor is μs = 0.80. The center of mass of the pencil is located 9.0 cm from the tip of the eraser and 11.0 cm from the tip of the pencil lead. Find the minimum angle θ for which the pencil does not slip.arrow_forwardA garage door is mounted on an overhead rail as shown below. The wheels at A and B have rusted so that they do not roll, but rather slide along the track. The coefficient of kinetic friction is 0.50. The distance between the wheels is 2.00 m, and each are 0.50 from the vertical sides of the door. The door is uniform and weighs 977 N. It is pushed to the left at constant speed by an external horizontal force. If the distance h is 1.54 m: a. What is the vertical component of the force exerted on the wheel A by the track? b. What is the vertical component of the force exerted on the wheel B by the track? c. Find the maximum value h can have without causing one wheel to leave the track. A В K 2.00 m h k- 3.00 m Figure 3. The two wheel track of a rusted garage door.arrow_forwardIt is sometimes important to determine the location of a person’s center of gravity(which coincides with the center of mass because the gravitational force is uniform at the scale of the human body). This can be done with the arrangement shown intheFigure below. A light (negligible mass) plank rests on two scales that read Fg1=380 N and Fg2=320 N. The scales are separated by a distance of 2.00 m. How far from the woman’s feet is her center of gravity?arrow_forward
- Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles in the back of the neck must therefore exert a force to keep it erect. That is why your head falls forward when you fall asleep in class. If the perpendicular distance between the line of action for the weight of the head and the pivot point is rW⊥ = 2.0 cmand the perpendicular distance between the line of action for the force the muscles exert on the head and the pivot point is rM⊥ = 4.6 cm, determine each of the following. (Assume the weight of the head is 50 N.) (a) the force exerted by these muscles(b) the force on the point of supportarrow_forwardA 3000-lb car is parked on a 30° slope, facing uphill. The center of mass of the car is halfway between the front and rear wheels and is 2 ft above the ground. The wheels are 8 ft apart. Find the normal force exerted by the road on the front wheels and on the rear wheels. 0 = 30° l = 2 ft 2 = 4 ft mg Hints : The car is in stationary equilibrium, so : 1) the total force on the car must be 0 (i.e., along the incline and perpendicular to it), 2) the total torque about any point must be 0. (choose as point of rotation the center of mass) Remember friction is: f = µ N Three equations , three unknowns. Solve for N1 and N2.arrow_forwardA 68 kg man is in the prone position while during pushups. His feet and hands are 95 cm and 42 cm away respectively from his center of mass. What is the normal force exerted by on each (a) hand and (b) foot?arrow_forward
- A 10.0-kg monkey climbs auniform ladder with weight w =1.20 x 102 N and length L =3.00 mas shown in Figure P8.94. The ladderrests against the wall at an angleof θ = 60.0°. The upper and lowerends of the ladder rest on frictionlesssurfaces, with the lower endfastened to the wall by a horizontalrope that is frayed and that can support a maximum tension of only 80.0 N. (a) Draw aforce diagram for the ladder. (b) Find the normal forceexerted by the bottom of the ladder. (c) Find the tensionin the rope when the monkey is two-thirds of the way up theladder. (d) Find the maximum distance d that the monkeycan climb up the ladder before the rope breaks. (e) If thehorizontal surface were rough and the rope were removed,how would your analysis of the problem be changed andwhat other information would you need to answer parts (c)and (d)?arrow_forwardA 1 490-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the centerline at a point 1.05 m behind the front axle. Find the force exerted by the ground on each front wheel (kN) Find the force exerted by the ground on each rear wheel (kN)arrow_forwardThe device shown in the figure below is one version of a Russell traction apparatus. It has two functions: to support the injured leg horizontally and at the same time provide a horizontal traction force on it. This can be done by adjusting the weight W and the angle θ. For this patient, his leg (including his foot) is 90.0 cm long (measured from the hip joint) and has a mass of 11.4 kg. Its center of mass is 41.0 cm from the hip joint. A support strap is attached to the patient's ankle 13.0 cm from the bottom of his foot. 1) What weight W is needed to support the leg horizontally? Express your answer in newtons. 2) If the therapist specifies that the traction force must be 16.0 N horizontally, what must be the angle θ? Express your answer in degrees. 3) What is the greatest traction force that this apparatus could supply to this patient's leg? Express your answer in newtons. 4) What is θ in that case? Express your answer in degrees.arrow_forward
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