One end of a uniform meter stick is placed against a vertical wall (Figure 1) .

Question

One end of a uniform meter stick is placed against a vertical wall (Figure 1) . 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 0.35.

Part A (I solved this)

What is the maximum value the angle ? can have if the stick is to remain in equilibrium?

The answer is 19 degrees

Part B

Let the angle between the cord and the stick is ? = 16 ?. A block of the same weight as the meter stick is suspended from the stick, as shown at (Figure 2) , at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium?

Express your answer using two significant figures.

Part C

When ? = 16 ?, how large must the coefficient of static friction be so that the block can be attached 14 cm from the left end of the stick without causing it to slip?

Express your answer using two significant figures.

Explanation / Answer

Since you have mentioned that you solved part A, so skipping that.

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=0.35

force of friction =f =mu*normal reaction=0.35*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

[ 0.35 T cosO ] L=mgL/2 +mg (L-x)

For translatory equilibrium,

total upward force= downward force

f + TsinO=mg + mg

0.35TcosO + TsinO=2mg

sinO=sin16=0.275

cosO= cos16= 0.961

For translatory equilibrium, total upward force= downward force

f + TsinO=mg+mg

0.3365 T+ 0.275 T= 2mg

0.6115 T= 2mg

mg=0.30575 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

0.3365 T=(1-x)mg+(1/2) mg

0.3365 T= [ (3/2) -x ]0.30575 T

x=0.399 m

the minimum value of 'x' (distance from wall) for which the stick will remain in equilibrium is 0.399 m = 0.40 m
______________________________________...

C )sinO=sin16=0.275

cosO= cos16= 0.961

Hence frction = f= muTcosO=mu*0.961T

For translatory equilibrium,
total upward force= downward force

f + TsinO=mg+mg

mu*0.961T+ 0.275T=2mg

mg=[ mu*0.961T+ 0.275T ] /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=14 cm =0.14 m

mu*0.961T=(0.86)mg+(1/2) mg

mu*0.961 T=1.36 mg

mu*0.961 T=1.36 *[ mu*0.961 T+ 0.275T ] /2

2mu*0.961 T=1.36[ mu*0.961 T+ 0.275T ]

2mu*0.961 T -1.36[ mu*0.961 T ] =1.35* 0.275T ]

mu* 0.961T (2-1.36)=1.36*0.275 T

mu = 0.608 = 0.61

When theta is 16 degree, the coefficient of static friction ( for block at 14 cm from wall) should be 0.608 (0.61)

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