### Educational Content on Jib Crane Analysis**Problem Statement:**After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure. The horizontal steel beam had a mass of 83.40 kg per meter of length, and the tension in the cable was \( T = 12430 \, \text{N} \). The crane was rated for a maximum load of 500 kg. If \( d = 5.580 \, \text{m} \), \( s = 0.594 \, \text{m} \), \( x = 1.650 \, \text{m} \), and \( h = 1.890 \, \text{m} \), what was the magnitude of \( W_L \) (the load on the crane) before the collapse? The acceleration due to gravity is \( g = 9.810 \, \text{m/s}^2 \).**Diagram Explanation:**The diagram depicts a jib crane with a horizontal beam supported by a cable. The beam has several labeled dimensions and angles:- \( d \): Horizontal distance from the pivot point P to where the load \( W_L \) is applied at the crane's end.- \( s \): Vertical distance from the top of the crane post to the attachment point of the cable.- \( x \): Horizontal distance between the crane post and where the cable meets the beam.- \( h \): Vertical height from the base to the attachment point P.**Incorrect Calculations:**1. **Magnitude of Load \( W_L \):** - Calculated value: \( W_L = 5008.9 \, \text{N} \) - Status: Incorrect2. **Magnitude of Force \( F_P \) at the Attachment Point P:** - Calculated value: \( F_p = 1270.25587 \, \text{N} \) - Status: IncorrectThe task involves recalculating these values considering the dimensions, forces, and gravitational acceleration provided to determine the true cause of the jib crane failure.

Given: The tension in the cable is 12430 N. The maximum allowable load is 500 kg. The mass per unit length of the horizontal beam is 83.40 kg. The length is 5.580 m. The distance s is 0.594 m. The distance x is 1.650 m. The height h is 1.890 m.Calculation: Calculate the mass of the horizontal beam. M=83.40 kg/m5.580 m=465.37 kg Write the expression for angle. tanθ=hd-s Substitute the values. tanθ=1.8905.580-0.594θ=tan-10.379θ=20.75° Write the expression for torque balance. Tsinθd-s-WLd-x-Mgd2=0 Substitute the values. 12430 Nsin20.755.580 m-0.594 m-WL5.580 m-1.1650 m-465.37 kg9.8 m/s25.580 m2=0WL5.580 m-1.1650 m=9233.38 N.mWL=2091.36 N Thus, the load on the crane before it collapsed was 2091.36 N.Write the expression for the net horizontal component of the force at point P. Px=Tcosθ Substitute the values. Px=12430 Ncos20.75=11623.73 N Write the expression for the net vertical force on the horizontal beam. Py=WL+Mg-Tsinθ Substitute the values. Py=2091.36 N+465.37 kg9.8 m/s2-12430 Nsin20.75=2248.15 N Write the expression for the resultant. FP=Px2+Py2 Substitute the values. FP=11623.73 N2+2248.15 N2=11839.14 N Thus, the net force on the point P is 11839.14 N.

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After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure.The horizontal steel beam had a mass of 83.40 kg per meter of length, and the tension in the cable was T = 12430 N. The crane was rated for a maximum load of 500 kg. If d = 5.580 m, s = 0.594 m, x = 1.650 m, and h = 1.890 m, what was the magnitude of WL (the load on the crane) before the collapse? The acceleration due to gravity is g = 9.810 m/s². W W = 5008.9 Incorrect What was the magnitude of force Fp at the attachment point P? Fp = 1270.25587 Incorrect

After an unfortunate accident occurred at a local warehouse, you were contracted to determine the cause. A jib crane collapsed and injured a worker. An image of this type of crane is shown in the figure.The horizontal steel beam had a mass of 83.40 kg per meter of length, and the tension in the cable was T = 12430 N. The crane was rated for a maximum load of 500 kg. If d = 5.580 m, s = 0.594 m, x = 1.650 m, and h = 1.890 m, what was the magnitude of WL (the load on the crane) before the collapse? The acceleration due to gravity is g = 9.810 m/s². W W = 5008.9 Incorrect What was the magnitude of force Fp at the attachment point P? Fp = 1270.25587 Incorrect

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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