### Problem 2.17: Moment Calculation for a Wheelbarrow**Problem Statement:**For the wheelbarrow shown, find the moment of the 100-pound (100#) weight about the center of the wheel. Also, determine the force \( P \) required to resist this moment.**Diagram Explanation:**The given image illustrates a side view of a wheelbarrow loaded with a weight of 100 pounds. Key details from the diagram include:- The wheelbarrow has two handles extending from the wheel axis.- The weight (100#) is positioned vertically above a specific point on the wheelbarrow.- The distance from the wheel (pivot point) to the point where the weight is applied is shown as 20 inches horizontally.- There are two segments indicating other distances on the handles: - 20 inches from the center of the wheel to the load application point. - Additional 20 inches to the end of the wheelbarrow handles where force \( P \) is applied.**Steps to Solve:**1. **Calculate the Moment due to the Weight:** - The moment \( M \) around the wheel’s center caused by the 100# weight can be calculated using the formula: \[ M = \text{Force} \times \text{Distance} \] Here: \[ M = 100 \text{ pounds} \times 20 \text{ inches} \] This gives: \[ M = 2000 \text{ pound-inches} \]2. **Determine the Force \( P \) to Resist the Moment:** - The resisting force \( P \) creates a counter-clockwise moment about the wheel’s center. - The distance from the pivot point to where \( P \) acts is 40 inches (20 inches + 20 inches). Using the moment equilibrium condition: \[ P \times 40 \text{ inches} = 2000 \text{ pound-inches} \] Solving for \( P \): \[ P = \frac{2000 \text{ pound-inches}}{40 \text{ inches}} = 50 \text{ pounds} \]**Conclusion:**- The moment of the 100# weight about the center of the wheel is 2000 pound-inches.- The force \(

moment is equal to force multiplied by perpendicular distance. moment about center of wheel =0

Answered: 2.17 For the wheelbarrow shown, find…

2.17 For the wheelbarrow shown, find the moment of the 100 weight about the center of the wheel. Also, determine the force P required to resist this moment.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A P = 10 kip 1.5 ft B C 2 ft The bars in the structure to the left are pinned and have circular cross-sections. The diameter of the cross- section is 1.5 inches. The bars are made of steel, which has an elastic modulus of 29,000 ksi. A point force of 10 kips is applied at point C, as shown. What is the change in length of member BC? Is the bar getting longer or shorter?
Determine the reactions at the supports. Assume the support at A is fixed and that at B is a roller. Take E = 29x103 ksi. The Moment of inertia of each segment is shown in the figure. (By Using The Force Method) A 9 ft- IAB 30 k = 600 in 18 ft B 30 k IBC= 300 in | C 12 ft-
A plate girder is held in a horizontal position by two cables as shown in FIGURE Q((a). Knowing that the weight of box at O is 30 kg, determine an equivalent resultant force and moment about point B when a = 40º.
The resultant "R" of the four forces A, B, C, and D is shown. Find the value of the angle θ and the magnitude of C.
The crank OA is shown. If OA = 57 mm, 0 = 60°, and 8 = 65°, determine the moment of the force F of magnitude F = 29 N about point O. The moment is positive if counterclockwise, negative if clockwise. Answer: Mo = i N-m
Q1: - For the frame shown in figure 1 below is subjected to two forces ( P=20 kN and Q=30 kN). Neglecting the weights of the bars, determine all forces that acting on each of the members. Use the two-force principle wherever applicable. 2 m 2 m B. A 60° D. 60° 4 m 2 m
In the system shown, the rigid bar ABC is subjected to a load w = 0.5 kip/in. Bars 1, 2 and 3 are made of steel and are hinged at both ends. a) State whether the system is statically determinate or indeterminate and the degree of indeterminacy. b) Determine the force in steel bars 1, 2, and 3 (indicating whether they are in tension or compression). c) Determine the displacement of point A. All steel bars have area A=1.25 in^2 L= 6 in Eac = 29×10^3 ksi
Determine the force in members BC, GA, and BG of the truss shown in the figure below. Let P = 175 kN. NOTE: If your answer is a compressive force then just put a negative sign. If your answer is a tensile force then just type the answer directly. (ALSO, DON'T FORGET THE UNITS) 2P 2P 30 60V60° 60°V60° C 30 D 3P 4m 4m 4m Solve the force in member BC. Solve the force in member GA. Solve the force in member BG.
Rod OA rotates about O in a horizontal plane. The motion of the 0.5-lb collar B is defined by the relations r= 10 + 6 cost and 0=(4-8), where r is expressed in inches, r in seconds, and 8 in radians. Determine the radial and transverse components of the force exerted on the collar when (a)1=0, (b) t=0.5s.
Reduce the given loading system to a force-couple system at point A. Then determine the distance x to the right of point A at which the resultant of the three forces acts. A Answers: 8" V R= i M= 320 lb X= i Force-couple system at A. The force is positive if up, and the moment is positive if counterclockwise. Resultant i 15" 24" 600 lb 350 lb lb lb-in. in.
A rod has a radius of 4m. Determine the smallest force F that must be applied along the rope in order to cause the curved rod to fail at the support A. This requires a moment of M=1400 N?m to be developed at A. 6 m B 45° 4 m F A 4 m 6 m
3m 1.5 m ļ ·x E A 2m D B -2m F 300 N C F₁ Consider the frame shown in figure 1. The load F1-680 N is applied to the pin. Determine the x and y components of force which pin C exerts on member ABC using scalar notation.