# Tough Mechanics A-Level Question

• May 7th 2008, 01:12 PM
helpishere
Tough Mechanics A-Level Question
A uniform metal bar AB of mass 80kg and length 16m.
It rests horizontally in equilibrium on two smooth supports N and M
N is 6m from A
M is at B

i)By taking moments about B, find the reaction at N
ii)Find the reaction at M

Any help greatly appreciated
• May 7th 2008, 01:35 PM
part i)

weight of the bar = mg = 9.81(80) = 784.8 N
It can be assumed that this weight acts at the center of the bar(8 m away from A or B)

$
R_N(x) - mg(8) = 0
$

where x = distance of normal force acting on the bar acting upwards at N from B = 16 - 6 = 10m
$R_N$ = reaction at N

therefore,
$
R_N(x) - mg(8) = 0
$

$
R_N(10) - mg(8) = 0
$

$
R_N= \frac {mg(8)}{10}
$

$
R_N= \frac {784.8(8)}{10}
$

$
R_N= 627.84N
$

part(ii)
resolving forces vertically:

$
R_N + R_M = mg
$

where R_M = reaction force action on the bar at M
$
R_N + R_M = mg
$

$
627.84 + R_M = 784.8
$

$
R_M = 784.8 - 627.84
$

$
R_M = 156.96 N
$
• May 7th 2008, 01:43 PM
topsquark
Quote:

Originally Posted by helpishere
A uniform metal bar AB of mass 80kg and length 16m.
It rests horizontally in equilibrium on two smooth supports N and M
N is 6m from A
M is at B

i)By taking moments about B, find the reaction at N
ii)Find the reaction at M

Any help greatly appreciated

Call the reaction forces N and M respectively.

Then from Newton's 2nd in the upward direction we have
$\sum F = -w + N + M = 0$
where w is the weight of the bar.

We have two unknowns in this equation so we need one more equation. Since the bar is not rotating we know that the sum of the torques about any axis is 0 Nm. So let's find the sum of the torques about the point B.
$\sum \tau = (8~m)w - (10~m)N = 0$
(Taking CCW to be a positive rotation.)

Solve this for N, then you can use the Newton's Law equation to find M.

-Dan