Abox is pushed down at an angle of 32 degrees on a rough surface. the box moves to the right. what equation should be used to find the net force in the y-direction? answers: a. fnet, y = fn - fgb. fnet, y= fp - ffc. fnet, y = fn-fg- fpcos(32)d. fnet, y = fn-fg-fpsin(32)
We have four forces acting on the box. Taking the upward direction as positive direction for the vertical axis:
- The push, which pushes down along the ramp inclined at 32 degrees. Its vertical component will be , where F is the magnitude of the push
- The weight of the box, of magnitude , with m being the mass of the box and g being the acceleration of gravity (9.8 m/s^2). This force is already directed vertically downward, so we don't need to resolve it into the vertical direction
- The frictional force, of magnitude , where is the coefficient of friction, directed upward along the inclined plane. Its vertical projection will be
- The normal reaction, whose magnitude is equal to the component of the weight perpendicular to the inclined plane, but with opposite direction (upward): . If we resolve it along the vertical direction, we get
So, the net force along the y-direction will be
find the surface area of the sphere to the nearest square unit. use a calculator.
a. 254 in.2
b. 64 in.2
c. 127 in.2
d. 1,018 in.2
Where I have assumed the 32 degrees was measured down from the horizontal so the sine would give the vertical component.