- #1
isukatphysics69
- 453
- 8
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?