A 200 lbf force, F1, is applied to the right end of a 30 ft lever at point A. The weight of the lever is 150 lbf and can be modeled as a vertical force, W, at the center of the lever. The applied moment at O is 2500ft-lbf in the clockwise direction. Determine the magnitude and direction of the resultant force, and its location with respect to point A. Note: This problem is covering the Day 1 material from Week 7. Do not confuse Resultant forces with Reaction Forces. Reaction forces were covered during Day 2 and will be on HW 8.1. 30 ft M. 60° W F

The screw-eye shown below experiences two forces. +ve Y axis 10⁰ Direction: 0= X KN KN YN Given that X = 13 kN and Y = 500 N calculate the magnitude and direction of the resultant force. Enter your answer for the magnitude in kN and the direction in degrees with respect to the positive x-axis, both correct to three significant figures in the appropriate boxes below. Magnitude: R= degrees 15° +ve X axis

Replace the loading (F1 and F2) by an equivalent resultant force and couple moment at point A. Enter your responses in the blanks provided. Upload your workings. What is the magnitude of the resultant force? Enter your response in Newtons to three significant figures.

Review Determine the moment of force F about the y axis. Determine the moment of force F about the , the y, and the z axes. Assume F = 570 N. (Figure 1) Express your answer to three significant figures and include the appropriate units. My = -137 N-m Submit Previous Answers v Correct Correct answer is shown. Your answer -136.8 N-m was either rounded differently or used a different number of significant figures than required for this part. Figure < 1 of 1 Part C Determine the moment of force F about the z axis. Express your answer to three significant figures and include the appropriate units. ? M, = Value Units 3 m Submit Request Answer 2 m 2 m