Question

A 2 kg rock is suspended by a massless

string from one end of a 7 m measuring stick.
What is the weight of the measuring stick
if it is balanced by a support force at the
1 m mark? The acceleration of gravity is
9.81 m/s^2
Answer in units of N.

Answer 1

Weight = 7.848 N

Step-by-step explanation:

Given:

Mass of the rock (m) = 2 kg

Length of the stick (L) = 7 m

Distance of support from the rock (x) = 1 m

The weight of the stick acts at the center, which is at a distance of 3.5 m from one end.

So, let 'M' be mass of the measuring stick.

Distance of 'W' from the supporting force (d) = 3.5 - 1 = 2.5 m

Now, for equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments.

Sum of clockwise moments = Sum of anticlockwise moments

`m* g* x=M* g*d\\\\M=(mx)/(d)`

Plug in the given values and solve for 'M'. This gives,

`M=(2* 1)/(2.5)=0.8\ kg`

Now, weight of the stick is given as:

`W=Mg=0.8* 9.81=7.848\ N`

Answer 2

The weight at
`1`m mark is 7.848 N.

Step-by-step explanation:

We have been given,

Mass of the rock=
`2` kg

Acceleration (gravity) =
`9.8` m/s^2

Length of the string =
`7` m

To find the weight at
`1` m mark let it be W.

So,

We know that torque will be balanced here.

Moment arm for the force (W) is
`3.5 - 1=2.5` m

As it is of
`7` m length so
`3.5` m is the COM (center of mass) of the string.

Let the moment be
`L_1\ and\ L_2` on clockwise and anti-clockwise direction.

`L_1=L_2`

`2.5*W = 2* 9.81*1`

Dividing both sides with
`2.5`

`W=(2*9.81*1)/(2.5) =7.848\N`

So the weight is 7.848 N.