## Homework Statement

A mass of 8 kg lies on a frictionless, horizontal floor. A force of 145 Newtons is applied to the mass at an angle above the positive x axis and a force of 16 Newtons is applied to the mass at an angle of 24.5 degrees below the negative x direction. If the angle for 145 Newtons is increased until the object just starts to leave the surface, what is the magnitude of the acceleration on the mass in m/s2 at this point?

## The Attempt at a Solution

F1=145 N with unknown angle
F2=16 N with angle of 24.5 degrees
F=ma
145N=8kg(a)
a=18.125

Is this correct?

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The solution is incorrect. wouldn't you suspect anything when the question gives so many variables and yet you only used half of them? :tongue2:

First, what makes the object leave the floor?

Its the vertical force! So resolve everything in vertical form...
Force upward is equal force downward at point of leaving surface.

145sinθ=16sin(24.5)+8(9.8) <---assume gravity is 9.8

then you will get θ=35.9° (3 s.f)

Using the angle, you can find the acceleration, by resolving horizontal forces.