A2 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.

The weight at m mark is 7.848 N.

Explanation:

We have been given,

Mass of the rock= kg

Acceleration (gravity) = m/s^2

Length of the string = m

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

So,

We know that torque will be balanced here.

Moment arm for the force (W) is m

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

Let the moment be on clockwise and anti-clockwise direction.

Dividing both sides with

So the weight is 7.848 N.

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

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

Now, weight of the stick is given as: