Q1:A diving board of length 3.00 m is supported at a point 1.00 m from the end,

Question

Q1:A diving board of length 3.00 m is supported at a point 1.00 m from the end, and a diver weighing 460 N stands at the free end (Figure 1) . The diving board is of uniform cross section and weighs 265 N .

a.Find the magnitude of the force at the support point.

b.Find the direction of the force at the support point.

c.Find the magnitude of the force at the left-hand end.

d.Find the direction of the force at the left-hand end

Q2:Two forces equal in magnitude and opposite in direction, acting on an object at two different points, form what is called a couple. Two antiparallel forces with equal magnitudes F1=F2=7.00 N are applied to a rod as shown in the figure (Figure 1) .

a.What should the distance l between the forces be if they are to provide a net torque of 5.00 Nm about the left end of the rod?

b.Is the sense of this torque clockwise or counterclockwise?

c.Repeat part A for a pivot at the point on the rod where F 2 is applied.

d.Repeat part B for a pivot at the point on the rod where F 2 is applied.

Explanation / Answer

answer 1 :-

Given that the system is in equilibrium the clock-wise and anti clock-wise moments about ANY point in the system are equal.
• Board weight acts down from centre of board

a) Consider the left hand end as a pivot point (any force / reaction at a pivot has no moment in the system)

C-W moments = A C-W moments
(460kg x 3m) + (265kg x 1.5m) = (Fs x 1m)
.. Fs = 1777.5 N (up)
b) up

c) Consider the support point as a pivot ..
C-W mom. = A C-W mom.
(460kg x 2m) + (265kg x 0.5m) = (F x 1m)
.. F = 1052.5 N (down)

d) down

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Answer 2.

a)     l = torque / force

          = 5 N.m / 7 N

          = 0.7143 m

          = 71.43 cm

b) clockwise

c) l = torque / force

          = 5 N.m / 7 N

          = 0.7143 m

          = 71.43 cm

d) clockwise

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