Chapter 5, Problem 6P

Please provide all writeen work and FBD. Thank you

draw free body diagram and A and B please. picture is pretty clear! Block is labeled A (under A is 25 degrees) above is 45 degrees and F pulling down is 50 degrees

solve show all steps, no copied answers

10-23 Problem 10-23 Clamping fixture. A holding fixture for a workpiece 37.5 mm thick at the clamp locations is being designed. The detail of one of the clamps is shown in the figure. A spring is required to drive the clamp upward when removing the workpiece with a starting force of 45 N. The clamp screw has an M10 X 1.25 thread. Allow a diametral clearance of 1.25 mm between it and the uncompressed spring. It is further specified that the free length of the spring should be Lo ≤ 48 mm, the solid height L, 31.5 mm, and the safety factor when closed solid should be n, 1.2. Starting with d = 2 mm, design a suitable helical coil compression spring for this fixture. For A227 HD steel, wire diameters are available in 0.2-mm increments between 0.2 to 3.2 mm. Clamp screw Spherical washer Slot Workpiece Clamp Groove Pin

1. Calculate the degree of freedom of the following mechanism shown in Fig. 4-1. If there are multiple pin joints, passive degree of freedom or redundant constraints, mark them in the figure. Fig. 4-1

Use the drawing below to answer the following two questions. a. Determine the equivalent spring stiffness of the following system. Simplify the expression (ie – single common denominator. Your answer should be an expression of just K’s – no numbers). b. If the weight on top is 20 lb and the K1 = K4 = 50 lb/in, K2 = 33 lb/in and K3 = K5 = 25 lb/in, what is the natural frequency of the system?

Could I please get some help with this problem please and thank you!

dont give wrong solution

In the six-bar mechanism in the figure, the joint points are shown with J1, J2,…., J7, and the limb numbers from 1 to 6 as shown in the figure. The limb lengths are as follows: ?1 = 1, ?2 = 0.3, ?3 = 0.7, ?4 = 0.8, ?5 = ?6 = ?8 = ?9 = 0.5, ?7 = 0.7 m A force of F = 1000 N and a moment of magnitude M = 500 Nm is applied to the same limb in the direction shown in the figure, right in the middle of the limb number 6 and perpendicular to the limb and in the direction shown in the figure. In this case, the mechanism is Calculate the moment Mt, which keeps it in balance. İ=6      q2=126    q3=28.10 q4=134.31                  q6=97.44 q7=-1.17 q8=74.32 q=angle