Position of Normal Force in a Pulled 2D Rectangular Block

AI Thread Summary
A 2D rectangular block weighing 403N is pulled at a constant speed with a force of 225N at an 18-degree angle, creating a torque that affects the normal force's position. The normal force is calculated as 333.471N after accounting for the vertical component of the applied force. To maintain equilibrium, the torque from the applied force must balance with the torque from friction, requiring the calculation of both the x and y components of the applied force. The net torque around the center of mass must equal zero, leading to an equation that allows for solving the distance of the normal force from the lower left corner. Understanding these forces and their interactions is crucial for determining the exact position of the normal force.
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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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