- #1

Nanabit

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I'm very very bad at summing up the forces. I know that the sum of all forces in both the x and y directions is zero, and the same with torque, but honestly I don't know where to start because I have no experience in this type of problem. Same with #2.

2. A vertical post with a square cross section is 11.0 m tall. Its bottom end is encased in a base 1.50 m tall, which is precisely square but slightly loose. A force 5.60 N to the right acts on the top of the post. The base maintains the post in equilibrium.

Find the force which the top of the right sidewall of the base exerts on the post. (to the left and to the right.) b. Find the force which the bottom of the left sidewall of the base exerts on the post. (to the left and to the right.)

3. A carpenter's square has the shape of an L (d1 = 16.0 cm, d2 = 4.00 cm, d3 = 4.00 cm, d4 = 11.0 cm). Locate its center of gravity. (Hint: Take (x,y) = (0,0) at the intersection of d1 and d4) (d1 is the length of the long part of the L, d2 is the width of it, d4 is the length of the bottom part of the L, and d3 is the width of it.)

I know what the formula is for this, but I am a bit confused because I don't have any masses. Plus, are the x value and y values in the formulas just the distances in the x and y directions??

4. For safety in climbing, a mountaineer uses a 45.0 m nylon rope that is 10.0 mm in diameter. When supporting the 75.0 kg climber on one end, the rope elongates by 1.80 m. Find Young's modulus for the rope material.

y = FLo/change in L A F= 75 kg(9.8 m/s^2); L = 45.0 m; change in L = 1.8 m; A = pi r^2 = 7.85x10^-5. What am I missing?

Thanks. Sorry I'm so clueless.