A uniform plank of length 5.1 m and weight 235 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 458 N walk on the overhanging part of the plank before it just begins to tip?

Question

Drawing:

0 (person)
---------------------
^ ^ (d = 1.1 from lever to edge)

We just learned torque today and this was assigned as homework...well I thought that since torque is constant....i just calculated the torque of the plank and then divided that by the mass of the person..but this is wrong..what should i do?

Answer 1

Huh? You need to work on the conditions of equilibrium:
1) the sum of the vertical forces is zero ( weightplank + weight man + rightsupport force + left support force) Make the support forces up, weight down. If you are wrong, you get a negative).
2) The sum of all torques about any point is zero. Choose the point on one of the forces, either support, or in the middle...it makes the equations simpler.

Now you have two equations, and three unknowns: x, and the two support forces.But the first condition lets you solve one support force in terms of the other, so you have two equations, two unknowns. Solve for x. There is a bit of algebra, so be neat.

Answer 2

http://physicsforums.com/archive/index.php/t-98650.html

That is a discussion of a similar problem.

Answer 3

Consider the torque about the support that is 1.1 m from the end. If the plank is just starting to tip, there is no force at the other support, and the net torque is zero. Let X be the distance from the right support where the person is standing. The plank weight acts through the Center of Gravity in the middle of the plank, which is 1.45 m left of the right support.

Applying a torque balance,
1.45 * 235 = 458 * X

Solve for X

Answer 4

thank you very much!!!