It's my first time posting here; I'm usually able to get help from my teacher and friends at school, but I'm in a bind: I was absent for most of the torque/statics unit this week and I have a test tomorrow that can't be re-scheduled. Here's a few problems similar to the ones I'm to be tested on; if I could be guided through these I'd be eternally grateful.(adsbygoogle = window.adsbygoogle || []).push({});

1a. The problem statement, all variables and given/known data

A brick is placed on a bike pedal to provide torque and cause the bike to roll. If the brick weighs 8kg and the pedal is 16cm away from the crankshaft's axis of rotation, find the torque applied to the crankshaft when the pedal is in each of the following positions: straight forward, 45 degrees towards the front and down, and straight down.

1b. Relevant equations

t = m g x sin( theta) (is this right?)

1c. The attempt at a solution

Straight down-zero (no distance between the center of rotation and the point at which the weight acts)

at 45 degrees t = 8 * 9.8 * 0.16 * sin(45deg) = 8.87 Nm

and at 90 t = 8 * 9.8 * 0.16 * 1 = 12.5 Nm

Really not sure if these are right...but seems somewhat logical...

2a. The problem statement, all variables and given/known data

A basic draw bridge over a river is operated by two chains, one attached to each upper corner of the bridge. The bridge's mass is 300kg and is stationary at a 55 degree angle upwards from the horizontal. At this angle the chains are perfectly horizontal. Find the tension in each chain, the force exerted by each of the hinges on the bridge (assuming there are two identical hinges, one at each bottom corner), and the tension in each chain required to begin lifting the bridge from a far/open position.

2b. Relevant equations

See below...it's a start...

2c. The attempt at a solution

I don't have the length of the chain, or the door? Does this matter?

Assuming the length of the door = 1, the height of the chain, when horizontal (with a 55degree right traingle) = cos 35 = .819

The new triangle, when the door is horizontal is A (door) = 1, B (height) = .819

the angle between the chain and the door = invtan(.819/1) = 39.3175

I know the weight on the door = 300kg, acting at .5 so the vertical component of the force on the two chains combined = 150kg.

So the total tension on the two chains = the vertical force (150kg) / sin 39.3175 = 150kg / .6336 = 236.736kg

Each chain then has a total tension of 236.736/2 = 118.368

Not really sure to go from here?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Torque/statics problems

**Physics Forums | Science Articles, Homework Help, Discussion**