# Thread: Tough Mechanics A-Level Question

1. ## 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

2. 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)

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

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

therefore,
$\displaystyle R_N(x) - mg(8) = 0$
$\displaystyle R_N(10) - mg(8) = 0$
$\displaystyle R_N= \frac {mg(8)}{10}$
$\displaystyle R_N= \frac {784.8(8)}{10}$
$\displaystyle R_N= 627.84N$

part(ii)
resolving forces vertically:

$\displaystyle R_N + R_M = mg$

where R_M = reaction force action on the bar at M
$\displaystyle R_N + R_M = mg$
$\displaystyle 627.84 + R_M = 784.8$
$\displaystyle R_M = 784.8 - 627.84$
$\displaystyle R_M = 156.96 N$

3. 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
$\displaystyle \sum F = -w + N + M = 0$
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.
$\displaystyle \sum \tau = (8~m)w - (10~m)N = 0$
(Taking CCW to be a positive rotation.)

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

-Dan