Problems involving "diagonal" forces
Let's begin by examinine a force **F** pulling up at an angle **θ** on one side of an object which is located on a rough horizontal plane. A freebody diagram of this situation is shown below.
Our first step will be to resolve, or break down, the diagonal force into its horizontal and vertical components. These components, **
F**_{x} and **
F**_{y} are shown in the next diagram.
Since the object is in vertical equilibrium, the relationship ΣF_{y} = 0 will produce the equation
N = mg - F sin θ
If **F** has been pushing down at angle **θ** on the object instead of pulling up at angle **θ**,
then the value for the normal would change to N = mg + F sin θ.
If the object is also in horizontal equilibrium, that is, it is either at rest or moving with a constant velocity towards the right, the relationship ΣF_{x} = 0 yields the equation
f = F cos θ where f = µN.
If the object were accelerating to the right, the equation that could be used to calculate the object's horizontal acceleration would be
net F_{x} = ma
T cos θ - f = ma where f = µN.
Notice that the magnitude of the acceleration will be changed by the value of **f** since it is impacted by the value of the normal. Hence, it is easier to accelerate an object by pulling up on it (decreasing the normal force) than by pushing down on it (increasing the normal force).
Let's work some additional examples. |