Coefficient of friction for the block to be in equlibrium

[kishlaysingh04]

Homework Statement

a square box held with a person's fingers pushing downward one quarter of the way along the top of the box and with their thumb pushing side wards on the bottom of the side face of the box. Friction between the person's fingers and the box is obviously very important if the box is not to fall.Determine the required coefficient of friction to keep the box from falling.

θ=tan-1 (1/2) and α= 45°

2. The attempt at a solution

I have drawn he following forces to show you

the equations which i have got is:

μN₂= N₂ + Mg

and by balancing the torque I have got :

N₂ x root 5a x sinθ = μN₂ x 2 root 2 x cosα[

 

[kishlaysingh04]

can anyone help me??

 

[andrevdh]

The forces also cancel in the x-direction.
The torque is the product of the force and its lever arm,
which is the perpendicular distance to its line of action.

 

[ehild]

the equations which i have got is:

μN₂= N₂ + Mg

It is μN₂=N₁+Mg

and by balancing the torque I have got :

N₂ x root 5a x sinθ = μN₂ x 2 root 2 x cosα[

The torque equation is not correct. Multiply each force by the distance of the line of the force from the CM. ehild

 

[HalfLostAndSearching]

I would first clarify the assumptions and variables used in this problem. It seems that the box is in equilibrium, meaning that all forces are balanced and there is no net acceleration. The person's fingers are applying a downward force (N2) and a sideways force (N1), while the weight of the box (Mg) is acting downwards. The coefficient of friction (μ) is the ratio of the frictional force to the normal force, and it is a measure of how easily the box slides on the surface.

Based on the equations provided, it appears that the person's fingers are pushing down with a force equal to half of the weight of the box (N2 = Mg/2) and the angle of inclination (θ) is 26.6 degrees. The angle of the sideways force (α) is 45 degrees. To determine the required coefficient of friction, we can use the equation μN2= N2 + Mg, which states that the frictional force (μN2) must be equal to the sum of the downward force (N2) and the weight of the box (Mg).

However, this equation only takes into account the vertical forces. To also consider the torque, we can use the equation N2 x root 5a x sinθ = μN2 x 2 root 2 x cosα. This equation takes into account the distance from the point of rotation (a) and the angles of the forces. By solving these equations simultaneously, we can determine the required coefficient of friction to keep the box from falling.

It is important to note that the coefficient of friction can vary depending on the surface of the box and the surface it is resting on. It is also important to consider the precision and accuracy of the measurements and calculations used in this problem. Further experimentation and analysis may be necessary to determine the exact coefficient of friction in this scenario.