As shown below, a beam of length L is in equilibrium and is attached to a cable which is pulling it against a frictionless wall. The beam is not attached to the wall and the wall is only exerting a normal force on the beam. Additionally, there is a 130 N box hanging on a wire from the end of the beam. Determine the cable tension & the beam weight. 0.63 L 45° tension = 1.90 N beam weight 128.65 N %3D 130 N

As shown below, a beam of length L is in equilibrium and is attached to a cable which is pulling it against a frictionless wall. The beam is not attached to the wall and the wall is only exerting a normal force on the beam. Additionally, there is a 130 N box hanging on a wire from the end of the beam. Determine the cable tension & the beam weight. 0.63 L 45° tension = 1.90 N beam weight 128.65 N %3D 130 N

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter10: Virtual Work And Potential Energy

Section: Chapter Questions

Problem 10.55P: Solve Prob. 10.54 assuming that A and B are connected by a spring of stiffness k = 0.3 W/b and free...

See similar textbooks

Similar questions

please is urgent The 7.8m steel I-beam has a mass of 800kg and is lifted into position by applying a force P on the trolley to which the lower end B is secured. Cable AC is tied center of the beam and is horizontal when B is exactly below C. Determine, in function of x, the force P needed to maintain equilibrium from x=0 to the value of x at which A' lies exactly below C. Find the maximum value of P and the corresponding value of x.
A cable with supports at the same elevation has a span of 550 ft. The cable weighs 4 lb / ft and the sag at midspan is 100 ft. Determine The tension in the cable at midspan.
Determine the tensions T1 and T2 in the strings required to maintain equilibrium for the suspended object weighs W = 38.7N shown in the figure. Take, α = 29.8° & β = 50.2°. The value of Tension, T1 (N) = The value of Tension, T2 (N) =
A cable with supports at the same elevation has a span of 550 ft. The cable weighs 4 lb / ft and the sag at midspan is 100 ft. Determine The length of the cable.
Determine the force in the cable OC and the elongation of the spring OA, OB necessary to support a box of mass (M) 400 kg as shown in the figure below. If a = 4 m, b = 5 m, c = 3 m, k1= 5000 N/mm and k2= 4000 N/mm. All calculation using equilibrium of particles equation and must be in Newton (N) and with FBD
PROBLEM 1: The 1500-kg beam ABC has fixed support at A and connected to beam BD by hinge at B. Beam BD has a mass of 2000 kg and supported by a cable at D. The cable is attached to the 20-cm-diameter pulley at E and pulled vertically at the other end, by tension T. Given: a = 1.5 m; b = 0.9 m; c = 1.8 m d = 0.6 m; e = 1.4 m; θ = 75°. Calculate the required tension to maintain in equilibrium. (ANSWER: 12.830KN) Calculate the moment reaction at A. (ANSWER: 28.499) Calculate the total reaction at A. (ANSWER: 22.192) Calculate the vertical reaction at E. (ANSWER: 25.223)
Determine the tensions T1 and T2 in the strings required to maintain equilibrium for the suspended object weighs W = 38.7N shown in the figure. Take, α = 29.8° & β = 50.2°. The value of θ1 (Degree) = The value of θ2 (Degree) = The value of θ3 (Degree) =
Make the free body diagram and calculate the reaction efforts at the points with movement restrictions imposed. Data: L1 = 1.9m L2 = 3.3m L3 = 3.5m P1 = 75kN P2 = 140kN
Statics-Two Force Members 1. Determine the force P required to maintain equilibrium of the pulley system, and the reactions at the ceiling. 2. Assume W=1.25kN 3. Show and clearly label all Free Body Diagram to determine solution.
A cable with supports at the same elevation has a span of 550 ft. The cable weighs 4 lb / ft and the sag at midspan is 100 ft. Determine The tension in the cable at a support.
Draw the Free Body Diagram and solve the problem
The three-bar truss in Fig. a is subjected to a horizontal force of 5 kip. If the cross-sectional area of each member is 0.20 in2, determine the horizontal displacement at point B. E = 29(103) ksi.