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:

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

statement #3 on andre"s list is false.