2. Each block has a weight of 400 lb. Determine how far the force P can compress the spring until block B slips on block A. What is P for this to occur? The inclination between the block is 5-12 Equd Mct. Calcoloter k = 800 lb/ft μ's = 0.5 30° 5 B A 12 s=0.2

2. FORCE METHOD An overhang beam ABC is loaded as shown. Determine the reactions at the supports due to loading and a support settlement of 1 mm at the roller B. Take E = 200 GPa and I = 500(106) mmª. 20KN 20 k /m C 3m Im

The rigid bars AC and BCD are supported by pins at A and B and by a steel rod at D. There is a roller connection between the bars at C. Compute the vertical displacement of point D caused by the loading. Cross-sectional area of rod is 250 mm² and E = 200 GPa.

%3D If the intensity of the distributed load acting on the beam is w = 3KN/m, determine the reactions at the roller A and pin B. 3 TA

1. The beam is subjected to a load F = 400 N and is supported by the rope and the smooth surfaces at A and B. a. Draw the free-body diagram of the beam. b. What are the magnitudes of the reactions at A and B? 45% 1.2 m- -1.5 m- F 11 B 30⁰

Hinge B joins members AB and BC. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) Determine the reactions at fixed support A and roller C. (Enter forces in kilonewtons and moment in kN · m.) Ax = O Ay= ? MA= ? Cy=? Magnitude and direction

I can not solve this problem because there is no pin in the middle can you help?

d)Determine the internal torque (in N-m) in section B.

Determine the reactions at A and E if P = 740 N. What is the maximum value which P may have for static equilibrium? Neglect the weight of the structure compared with the applied loads. 4900 N 279 A B| | L--x 2.5 m E D 3.0 m 3.0 m P Ay =_N Ex = Ax =_N Pmax = %3D

Six forces are acting on the wedge-shaped block as shown. It is known that F₁ = F₁ F₂ = F₂ = 400 kN, F3 = F₂ = 600 kN, a = 10 m and b = 8 m. The six forces form three couples. Determine the resultant couple. (1). M = 4800kN-m (†) (3). M = 8264kN-m (†) F₁ 30⁰ a (2). M = 4800kN-m (↓) (4). M = 8264kN·m (↓) = 200 KN, F₂