A shop sign weighing 210 N hangs from the end of a uniform 180-N beam as shown in (Figure 1). Figure 35.0% -1.35 m- 1.70 m Paul's 。。 Auto Repair 1 of 1 > Part A Find the tension in the supporting wire (at 35.0°). Express your answer to three significant figures and include the appropriate units. Fr= 395 μA Submit Previous Answers Request Answer X Incorrect; Try Again ▾ Part B FH₂ = N Find the horizontal force exerted by the hinge on the beam at the wall. Express your answer to three significant figures and include the appropriate units. Enter positive value if the direction of the force is to the right and negative value if the direction of the force is to the left. μA Value ? Units P Pearson ?

The door has a weight of 230 lb and a center of gravity at G. (Bouto 1) Figure ▼ Part C NA = 1 of 1 HÅ Value Determine the constant force F that must be applied to the door to push it open 19.0 ft to the right in 5 s. starting from rest. Express your answer to three significant figures and include the appropriate units. Submit Request Answer Part B. Value Units Submit Previous Answers Request Answer X Incorrect: Try Again; 5 attempts remaining μA Also, find the vertical reaction at the roller B Express your answer to three significant figures and include the appropriate units. No - Value Submit Also, find the vertical reaction at the roller A. Express your answer to three significant figures and include the appropriate units. Previous Answers Request Answer Units Units ?

Reminders: Write out the equation you intend to use before substituting in it.

A diving board of length 3.00 m is supported at a point 1.00 m from the end, and a diver weighing 550 N stands at the free end (Figure 1). The diving board is of uniform cross section and weighs 270 N. For related problemealing ting and strateginn Figure K-1.00 m- -2.00 m 1 of 1 ▼ Part A Find the magnitude of the force at the support point. Express your answer in newtons. VG ΑΣΦ Fsupport Submit Part B = Request Answer ? N

2. Use the equilibrium conditions to find the weights w1, w2, and w; required to balance the mobile. Hint: start from the bottom rung and apply the conditions for equilibrium, then work your way up. 4 cm 8 cm 5 cm 1 cm W3 2 cm 4 cm w2 W1 0.5 N

The flanged steel cantilever beam with riveted bracket is subjected to the couple and two forces shown, and their effect on the design of the attachment at A must be determined. Replace the two forces and couple by an equivalent couple M and resultant Rat A. The couple is positive if counterclockwise, negative if clockwise. 2.01 KN 0.49 m 1.52 m- 76⁰ A 6 Answers: M= kN.m R=( i i+ i L-- 500 N·m 10.17 m 0.17 m 1.05 KN j) kN

1. A 160kg utility pole extends 12.0m above the ground. A horizontal force of 250N acts at its top and the pole is held in the vertical position by a cable. a. Draw a free-body (or force) diagram for the pole. b. Determine the tension in the cable. c. What are the reaction forces exerted at the lower end of the pole by the ground?   2. A 40.0kg uniform platform is suspended by two cables, 2.00m apart, where the centre of the platform lies midway between the cables. A 5.00kg paint bucket is placed on the platform, 24.0cm to the right of the right cable. Determine the tension in each cable.

A 12 000-N shark is supported by a rope attached to a 4.60-m rod that can pivot at the base. (a) Calculate the tension in the cable between the rod and the wall, assuming the cable is holding the system in the position shown in the figure. (Give you answer to three significant digits.) b) Find the horizontal force exerted on the base of the rod magnitude ------- N c) Find the vertical force exerted on the base of the rod. Ignore the weight of the rod. magnitude ------ N

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The flanged steel cantilever beam with riveted bracket is subjected to the couple and two forces shown, and their effect on the design of the attachment at A must be determined. Replace the two forces and couple by an equivalent couple M and resultant R at A. The couple is positive if counterclockwise, negative if clockwise. 1.97 KN 0.69 m 2.30 m 73⁰ A 7 Answers: M = i kN.m R=(i i+ i Y L- 450 N-m 1.34 KN 0.13 m 0.13 m j) KN