A horizontal force is applied to a
block that has a weight of 1.50 kilonewtons. The force holds the block at rest
on a plane inclined at 30.0 degrees above the horizontal. Assuming no friction, calculate the
normal force on the block.
We can call this normal force
capital 𝑁 and start out by drawing a diagram of the scenario. In this situation, our block sits
on a plane inclined at 30.0 degrees above the horizontal. In total, there are three forces
that act on the block. First, there’s its weight force
pulling it straight down. We’re told we apply a horizontal
force to the block we’ve called a capital 𝐹. And there’s also a normal force
acting on the block from the inclined plane.
We’re told the magnitude of the
weight force. But we don’t know 𝐹. And we want to solve for 𝑁. We are told that, under the
influence of these three forces, the block isn’t moving. It’s staying still in one place on
the plane. In our diagram, we can insert
coordinate axes to define positive and negative direction. We’ll say that positive motion in
the 𝑦-direction is perpendicular to the plane. And positive motion in the
𝑥-direction is up the plane parallel to it.
Our next task is to break up the
forces that are not already aligned completely with one of these axes directions
into components that are. First, we do that with our weight
force, breaking it up into components that are in the 𝑥-direction and in the
𝑦-direction. We can know that the 30.0-degree
angle we’ve been given, which we can call 𝜃, is equal to the angle in the
upper-left corner of this triangle. If we do the same breakdown of our
applied force 𝐹, we see it also can be divided into 𝑥- and 𝑦-components.
Our next step will involve writing
out force balance equations in the 𝑥- and in the 𝑦-directions. In the 𝑥-direction, the two forces
we have are 𝐹 times the cos of 𝜃 and negative 𝑊 times the sin of 𝜃. If we recall Newton’s second law of
motion, that the net force on an object is equal to its mass times its acceleration,
we can apply this to the forces in the 𝑥-direction of our scenario. We can write that 𝐹 cos 𝜃 minus
𝑊 sin 𝜃 equals the mass of our block multiplied by its acceleration in the
We know though that because the
block is motionless, this acceleration is zero. This means we can rewrite our
equation and know that 𝐹 times the cos of 𝜃 is equal to 𝑊 times the sin of
𝜃. Or dividing both sides by the cos
of 𝜃, 𝐹 is equal to 𝑊 times the tan of 𝜃. Both 𝑊, the weight force of the
block, and 𝜃, the angle of the inclined plane, are given to us. So we’ve effectively solved for 𝐹,
the applied force.
Without calculating it numerically,
we can move on to looking at forces in the 𝑦-direction. Applying Newton’s second law to the
forces in the 𝑦-direction, we can write that the normal force 𝑁 minus 𝑊 times the
cos of 𝜃 minus 𝐹 times the sin of 𝜃 is equal to the block’s mass multiplied by
its acceleration in the 𝑦-direction. Again, we know that that
acceleration is zero, which means that 𝑁, the normal force magnitude, is equal to
𝑊 cos 𝜃 plus 𝐹 sin 𝜃.
If we substitute in for 𝐹, 𝑊
times the tangent of 𝜃, we now have an expression for 𝑁, what we want to solve
for, in terms of known values 𝑊 and 𝜃. When we plug in 1.50 kilonewtons
for 𝑊 and 30.0 degrees for 𝜃 and calculate this expression, we find that 𝑁 is
1.73 times 10 to the third newtons. That’s the magnitude of the normal
force acting on the block.