80 lb 40 lb 331 33" A G - x 72" 40 lb B C 50 lb 30° 66"- 70 lb

B6 please no hand writing solution for other wise down vote

Please solve both part and correct answer otherwise dislike

A spring with an un-stretched length of 5 in expands from a length of 2 in to a length of 4 in. The work done on the spring is . in .lb. (A) k (2 in)2 (B) k [(4 in)? - (2 in)2] (C) - k [(4 in)? - (2 in)?] (D) k [(3 in)? - (1 in)*]

The figure shows a rigid bar that is supported by a pin at A and two rods, one made of steel and the other of bronze. Neglecting the weight of the bar, compute the force (in N) in the bronze rod caused by the 48512-Nload, using the following data:  length of steel = 1.42 m,   length of bronze = 2.36 m Area of steel = 639 mm,   Area of bronze = 319 mm Modulus of elasticity of steel = 180 Gpa, Modulus of elasticity of bronze = 81 Gpa  x = 0.58 m, y = 1.13 m, and z = 0.72 m. Round off the final answer to two decimal places.

Please answer all parts or don't answer bcz this is my last question of month

Use image below Read and Solve carefully please Write clearly and Box the Final answer(s) Express your answer to 3 Significant figures and include appropriate units. HINT : The elevator has a mass of 800 kg , the counterweight C has a mass of 200 kg, and the winding drum has a mass of 600 kg READ CAREFULLY

Solve the effective normal force on the rod CD Use units [N,mm] The mass of the load is: mk=59033kg

Find the relationship between 8₁, the elongation of rod (1), and the internal tensile force, F₁, in rod (1). Assume 8, is a positive value in units of mm and F₁ is a positive value in units of kN. So, the answer you enter for the ratio (8₁/F₁) must be in units of ( mm/kN). 8₁ F₁ = i ! mm/kN

B7