One end of a uniform meter stick is placed against a vertical wall. The other en
ID: 1482603 • Letter: O
Question
One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.50.
(a) What is the maximum value the angle can have if the stick is to remain in equilibrium?
(b) Let the angle be 20°. A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance x from the wall. What is the minimum value ofx for which the stick will remain in equilibrium?
(c) When = 20°, how large must the coefficient of static friction be so that the block can be attached 40 cm from the left end of the stick without causing it to slip?
Explanation / Answer
a) Let T be tension in the string. Resolving T into vertical and horizontal components,
Horizontal component TcosO presses meter stick towards wall
Vertical component isTsinO
normal reaction on stick due to wall =TcosO
coefficient of friction=mu=0.500
force of friction =f =mu*normal reaction=0.5*TcosO
As stick is in rotational equilibrium,net torque must be zero
Taking torque about the end held by cord ,
length of stick=L=1 meter
anticlockwise torque of friction = torque of weight
[ 0.5 T cosO ] L=mgL/2
mg=TcosO.......................(1)
For translatory equilibrium,
total upward force= downward force
f + TsinO=mg
0.5TcosO + TsinO=mg....................(2)
from (1) and (2)
TsinO=TcosO -0.5TcosO
cancelling T
sinO=0.5
The maximum value the angle ,if the stick is to remain in equilibrium, is 30.0 degree
b) Let T be tension in the string. Resolving T into vertical and horizontal components,
Horizontal component TcosO presses meter stick towards wall
Vertical component is TsinO
As weight of block is vertically downwards, normal reaction remains TcosO.
normal reaction on stick due to wall =TcosO
coefficient of friction=mu=0.500
force of friction =f =mu*normal reaction=0.5*TcosO
As stick is in rotational equilibrium,net torque must be zero
Assume wall is on left end of stick and cable is on right end
Taking torque about the end held by cord ,
length of stick=L=1 meter
suppose block of weight mg is at distance 'x' from wall, then
distance from cord is (L-x)
anticlockwise torque of friction = torque of weight+torque of wt of block
[ 0.5 T cosO ] L=mgL/2 +mg (L-x)
For translatory equilibrium,
total upward force= downward force
f + TsinO=mg + mg
0.4TcosO + TsinO=2mg
sinO=sin20=0.342
cosO= cos20= 0.9397
For translatory equilibrium, total upward force= downward force
f + TsinO=mg+mg
0.4699 T+ 0.342T= 2mg
0.8119 T= 2mg
mg=0.406 T
As stick is in rotational equilibrium,net torque must be zero
Taking torque about the end held by cord ,
fL=(L-x)mg+(L/2)mg
L=1meter
0.4699 T=(1-x)mg+(1/2) mg
0.4699 T= [ (3/2) -x ]0.406 T
x=0.3425 m
the minimum value of 'x' (distance from wall) for which the stick will remain in equilibrium is 0.3425 m
c ) sinO=sin20=0.342
cosO= cos20= 0.9397
Hence frction = f= muTcosO=mu*0.9397T
For translatory equilibrium,
total upward force= downward force
f + TsinO=mg+mg
mu*0.9397 T+0.342T=2mg
mg=[ mu*0.9397 T+ 0.342 T ] /2
As stick is in rotational equilibrium,net torque must be zero
Assume wall is on left end and cord is on right end of stick
Taking torque about the end held by cord ,
fL=(L-x)mg+(L/2)mg
L=1meter and x=40 cm =0.4 m
mu*0.9397 T=(0.60)mg+(1/2) mg
mu*0.9397 T=1.1 mg
mu*0.9397 T=1.1 *[ mu*0.9397 T+ 0.342 T ] /2
2mu*0.9397 T=1.1*[ mu*0.9397 T+ 0.342 T ]
2mu*0.9397 T -1.1[ mu*0.9397 T ] =1.1* 0.342 T
mu* 0.9397 T (2-1.1)=1.1*0.342 T
mu*0.8457 T = 1.1*0.342 T
mu=0.445
When theta is 18 degree, the coefficient of static friction ( for block at 15 cm from wall) should be 0.6748
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.