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Let's estimate he can push down with a force about a quarter of his weight. If he weighs 200 pounds, this would result in a force of 50 pounds, or 225 N. We also know that the force of friction (F) between his feet and the asphalt depends on the force with which he pushes down (N) and the "coefficient of kinetic friction"(μ) between the soles of his shoes, which we will assume are made of rubber, and the pavement. F = μN The μ between rubber and asphalt varies between 0.5 and 0.8. Let's assume a value of 0.7. Therefore, solving for stopping distance, we get: D = ½(2100kg)(18m/s)2/(0.7)(225N) = 2160 meters, or over 1.3 miles! The situation might be improved if he exerted his full 200 pounds, or 900 Newtons, of force against the ground. In that case: D = 1/2(2100kg)(18m/s)2/(0.7)(900N) = 540 meters (about a third of a mile) However, the amount of torque exerted on his ankles and knees might make that a problematic proposition.

All About G ForcesWhat's behind gravity forces, and how much of them can we take? By Peter TysonPosted 11.01.07NOVAA few summers ago I took my then nine-year-old daughter on a glider ride. Midway through, as we soared over a coastal landscape, I casually asked the pilot whether he could do any tricks. Without a word, he threw the plane into a dive. We were accelerating straight towards the ground. My daughter and I shouted and grabbed the armrests. Suddenly we were hit with that thrill-inducing pressure familiar from rollercoasters-tensed facial muscles, light-headedness, a sense of altered reality.

ingenious way to measure time of contact between tennis raquet and ball. 

Allows you to place Nodes & members in place to create a truss. It will also calculate the forces involved.

Force, Shape, Material,