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Box lifted off the ground by a tension force at a 32 degree angle

  • #1
PHYSICSROPEBOX.PNG
1. Homework Statement

The magnitude of the force F is slowly increased. The direction of the force remains the same. What is the magnitude of the force F at the moment the block looses contact with the floor?

Homework Equations


fnet=m*a[/B]


The Attempt at a Solution


fnety = tensionforcesin(32)-mg+fnormal
since box being lifted off ground fnormal = 0
since instantaneous acceleration is still 0 so 0=tensionforcesin(32)-24.01+0
24.01/sin32 = tensionforceydirection
tensionforceydirection = 45.31
plug this force into x component
fnetx = 45.31cos(32)
fnetx = 38.42
force mag = sqrt((38.42)^2+(45.31)^2))

incorrect answer
not sure what i am doing wrong here?
 

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Answers and Replies

  • #2
Orodruin
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0=tensionforcesin(32)-24.01+0
24.01/sin32 = tensionforceydirection
What happened here? In one equation you have the tension force (why don't you call it T or something more tangible?) and in the other its y-component.
 
  • #3
OK I JUST THOUGHT DEEPLY ABOUT IT AND REALIZED WHAT I DID WRONG AND GOT THE CORRECT ANSWER 45N
 
  • #4
What happened here? In one equation you have the tension force (why don't you call it T or something more tangible?) and in the other its y-component.
hi yes i realized what i was doing wrong the answer is 45N
 

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