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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.

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 for class=

Respuesta :

Answer:

Weight = 7.848 N

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

[tex]m\times g\times x=M\times g\timesd\\\\M=\frac{mx}{d}[/tex]

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

[tex]M=\frac{2\times 1}{2.5}=0.8\ kg[/tex]

Now, weight of the stick is given as:

[tex]W=Mg=0.8\times 9.81=7.848\ N[/tex]

Answer:

The weight at [tex]1[/tex]m mark is 7.848 N.

Explanation:

We have been given,

Mass of the rock= [tex]2[/tex] kg

Acceleration (gravity) = [tex]9.8[/tex] m/s^2

Length of the string = [tex]7[/tex] m

To find the weight at [tex]1[/tex] m mark let it be W.

So,

We know that torque will be balanced here.

Moment arm for the force (W) is [tex]3.5 - 1=2.5[/tex] m

As it is of [tex]7[/tex] m length so [tex]3.5[/tex] m is the COM (center of mass) of the string.

Let the moment be [tex]L_1\ and\ L_2[/tex] on clockwise and anti-clockwise direction.

[tex]L_1=L_2[/tex]

[tex]2.5*W = 2* 9.81*1[/tex]

Dividing both sides with [tex]2.5[/tex]

[tex]W=\frac{2*9.81*1}{2.5} =7.848\N[/tex]

So the weight is 7.848 N.

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