Concept explainers
A mobile is formed by supporting four metal butterflies of equal mass m from a string of length L. The points of support are evenly spaced a distance ℓ apart as shown in Figure P5.54. The string forms an angle θ1 with the ceiling at each endpoint. The center section of string is horizontal. (a) Find the tension in each section of string in terms of θ1, m, and g. (b) In terms of θ1, find the angle θ2 that the sections of string between the outside butterflies and the inside butterflies form with the horizontal. (c) Show that the distance D between the endpoints of the string is
Figure P5.54
Trending nowThis is a popular solution!
Chapter 5 Solutions
Physics for Scientists and Engineers
- A makeshift sign hangs by a wire that is extended over an ideal pulley and is wrapped around a large potted plant on the roof as shown in Figure P6.10. When first set up by the shopkeeper on a sunny and dry day, the sign and the pot are in equilibrium. Is it possible that the sign falls to the ground during a rainstorm while still remaining connected to the pot? What would have to be true for that to be possible? FIGURE P6.10 Problems 10 and 11.arrow_forwardAn early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary, the tension in the cable was 6500 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force. What was the tension in the cable when the craft was being lowered to the seafloor?arrow_forwardA 22 kg loudspeaker is suspended 2.4 m below the ceiling by two 2.70 m long cables that angle outward at equal angles. What is the tension in each of the cables?arrow_forward
- Ross and Rachel are fighting if they were on a break. Ross is trying to get his stuff weighing 150. N placed on an inclined plane by pulling it upwards parallel to the inclined plane with 150. N force. However, Rachel is trying to stop Ross by exerting 50.0 N perpendicular to the inclined plane. The coefficients of friction between the box and the incline are us = 0.450 and uk = 0.350. Rachel was able to stop Ross. The object is in static equilibrium. Rachel wins. The object is impending to move up. Nobody wins; the object slides down to the plane. Ross was able to get the object. The object slides up along the incline. 50 N F·R·I·E·N·D·S 30° Was Ross able to get the object or was Rachel able to stop him? (Ctrl)- 150 Narrow_forwardA mule is harnessed to a sled having a mass of 201 kg, including supplies. The mule must exert a force exceeding 1270 N at an angle of 30.3° (above the horizontal) in order to get the sled moving. Treat the sled as a point as a particle. a) Calculate the normal force (in N) on the sled when the magnitude of the applied force is 1270 N. (Enter the magnitude.) N (b) Find the coefficient of static friction between the sled and the ground beneath it. (c) Find the static friction force (in N) when the mule is exerting a force of 6.35 ✕ 102 N on the sled at the same angle. (Enter the magnitude.) Narrow_forwardHopeful for greater opportunities, Juan applies for an international scholarship. Fortunately, he gotaccepted and is now at the airport to travel to New Zealand. Waiting on queue at the entrance, Juanis holding his 30kg luggage up on a ramp that is inclined at an angle of 10° from the horizontal.He is holding on to his luggage at an angle of 65° from the ramp. Compute for the tension theluggage exerts on his hand and the normal force.arrow_forward
- A small box is held in place against a rough wall by someone pushing on it with a force directed upward at an angle of 260 above the horizontal. The coefficient of static and kinetic friction between the box and the wall are 0.40 and 0.30, respectively. The box slides down unless the applied force has a magnitude of 13 N. What is the mass of the box?arrow_forwardA block weighs 100. N and rests on a horizontal surface. If Ms = 0.50 and uk = 0.20 between the block and the horizontal surface, then (a) What is the magnitude of the frictional force acting on the block? Justify your What is the magnitude of the frictional force acting on the block if a horizontal force of 35.0 N is now used to push the block? Justify your answer with a calculation. What is the magnitude of the minimum horizontally applied force that will start the block sliding? Justify your answer with a calculation. If the magnitude of the applied horizontal force on the block is 55.0 N, what is the magnitude of the frictional force acting on the block? Justify your answer with a calculation.arrow_forwardA 4.00 kg box sits atop a 10.0 kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and the lower box and table is 0.600, while the coefficient of static frction between these same surfaces is 0.800. A horizontal pull of 150.0 N to the right is exerted on the lower box, and the boxes move together. What is the friction force on the upper box?arrow_forward
- A heavy sled is being pulled by two people, as shown in the figure. The coefficient of static friction between the sled and the ground is u, = 0.571, and the kinetic friction coefficient is y = 0.419. The combined mass of the sled and its load is m = 321 kg. The ropes are separated by an angle p = 25.0°, and they make an angle 0 = 29.7° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? minimum rope tension: N If this rope tension is maintained after the sled starts moving, what is the sled's acceleration? acceleration: m/s?arrow_forwardA 5N picture frame is supported by two strings that run from its upper corners to a nail on the wall. What is the tension on each string if each string makes an angle of 30° with the horizontal? * О 2.9 N O 5 N О 3.2 N 4.2 Narrow_forwardA block of mass 20kg is pushed against a vertical wall by force P. The coefficient of friction between the surface and the block is 0.2. If theta = 30 degrees, what is the minimum magnitude of P to hold the block still?I understand that in order for the block to sit motionless, the net forces acting on the block must be zero. I set my equation to be Net Force = 0 = Psin(theta) + Force Friction - Force Gravity.Which I rearranged as P = (Force Gravity - Force Friction)/sin(theta) or P = (mg-μ(mg))/sin(theta)Doing this gives me a value of 313.6N rather than 202.9N which I should be getting. What am I doing wrong?arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning