The diagram shows a stationary 5-kg uniform rod (AC). I m long, held against a w

Question

The diagram shows a stationary 5-kg uniform rod (AC). I m long, held against a wall rope (AE) and friction between the rod and the wall. To use a single equation to find force exerted on the rod by the rope at which point should you place the reference point tor computing torque? A B C D E An 800-N man stands halfway up a 5.0 m ladder of negligible weight. The base of the ladder is 3.0 m from the wall as shown. Assuming that the wall-ladder contact is frictionless, the wall pushes against the ladder with a force of: 150N 300 N 400 N 600 N 800 N A uniform 240-g meter stick can be balanced by a 240-g weight placed at the 100-cm mark if the fulcrum is placed at the point marked: 75 cm 60 cm L 50 cm 40 cm 80 cm

Explanation / Answer

37 )Ans C=C
=300
38 )Ans B
= Fw – mg( / 2 * cos ) = 0,

where = sum of torque,
Fw = force of wall against ladder =800
mg = man's weight * gravity (which is 800N)
= length of ladder =5m

The equation is equal to 0 because this is a static equilibrium problem,
where the ladder isn't moving & it's in equilbrium.

When you solve for Fw, the answer you get is 300N.

39) Ans A=75 cm

let the fulcrum be at x cm from (0 end)
lets take moments (weight*perpendicular distance) of weight at N >>

weight of N-0 section will act at P (x1=x/2)
weight of N-100 section will act at Q [x2=(100-x)/2]
weight of 240 g will act at 100 point [x3=(100-x)]
===============
[2.4*(x)]*(x/2) g = 240*(100-x) g + [2.4*(10-x)]*(10-x)/2*g

solve it X=75 cm

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