Position of Normal Force in a Pulled 2D Rectangular Block

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Homework Help Overview

The problem involves a 2D rectangular block being pulled at a constant speed, with specific dimensions and forces acting on it. The objective is to determine the distance from the lower left corner where the normal force acts, considering the applied force and its angle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the torque created by the applied force and the need to balance it with the torque due to friction. There are questions about the direction of the applied force and its components, as well as how to calculate the net torque to find the position of the normal force.

Discussion Status

Some participants have provided insights into resolving the applied force into its components and the implications for torque. There is ongoing exploration of how to set up the equations for torque and the relationship between the forces acting on the block. No consensus has been reached yet, and multiple interpretations of the problem are being considered.

Contextual Notes

Participants are working under the constraints of the problem statement, including the specific forces and angles given. There is an emphasis on maintaining equilibrium and considering the effects of friction, but some details remain unclear.

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Homework Statement


A 2D rectangular block is being pulled at a constant speed with a height of 2.2m and width of 51cm and weighs 403N. The pulling force of 225N is being applied at an 18 degree angle at a height of .666cm from the base of the box. What distance from the lower left corner does the normal force act?


Homework Equations


Normal force = 403 - 225sin18 = 333.471N
 
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The applied force will have a torque about the center of mass of the block. to balance this torque the position of normal force will be shifted accordingly. consider the torque due to friction if any.
 
So the torque of friction should equal the torque of the applied force in order to maintain equilibrium. So I need to find the torque of the x and y components of the applied force to find the applied force's torque and then divide that by the force of friction in order to get the distance.
 
Friction is a shear force. The net torque (sum of the moments) is zero, otherwise the block would rotate. Is the block being pulled left or right?

If the applied force is at an angle upward, then there is an upward component that lessens the normal force (e.g. weight downward).
 
I found the torque of the applied force which is 1.12268(225sin18)=78.05 but I just do not know where to go from here.
 
Resolve the applied force in horizontal (Fx) and vertical (Fy) direction.

Friction force is equal to the horizontal component but in opposite direction

the other forces are the normal force and the weight of the block.

consider the distance of the normal force equal to x from lower left corner.

Now calculate net torque due to all forces about any point (consider appropriate sign) and it should be zero. This will give the equation in x solve this to get x.
 

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