# Question Video: Using Newton's Laws to Determine Unknown Force on an Inclined Plane

A horizontal force is applied to a block that has a weight of 1.50 kN. 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.

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### Video Transcript

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 π 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, thereβs its weight force pulling it straight down. Weβre told we apply a horizontal force to the block weβve called a capital πΉ. And thereβs also a normal force acting on the block from the inclined plane.

Weβre told the magnitude of the weight force. But we donβt know πΉ. And we want to solve for π. We are told that, under the influence of these three forces, the block isnβt moving. Itβs staying still in one place on the plane. In our diagram, we can insert coordinate axes to define positive and negative direction. Weβll say that positive motion in the π¦-direction is perpendicular to the plane. And positive motion in the π₯-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 π₯-direction and in the π¦-direction. We can know that the 30.0-degree angle weβve been given, which we can call π, is equal to the angle in the upper-left corner of this triangle. If we do the same breakdown of our applied force πΉ, we see it also can be divided into π₯- and π¦-components.

Our next step will involve writing out force balance equations in the π₯- and in the π¦-directions. In the π₯-direction, the two forces we have are πΉ times the cos of π and negative π times the sin of π. If we recall Newtonβ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 π₯-direction of our scenario. We can write that πΉ cos π minus π sin π equals the mass of our block multiplied by its acceleration in the π₯-direction.

We know though that because the block is motionless, this acceleration is zero. This means we can rewrite our equation and know that πΉ times the cos of π is equal to π times the sin of π. Or dividing both sides by the cos of π, πΉ is equal to π times the tan of π. Both π, the weight force of the block, and π, the angle of the inclined plane, are given to us. So weβve effectively solved for πΉ, the applied force.

Without calculating it numerically, we can move on to looking at forces in the π¦-direction. Applying Newtonβs second law to the forces in the π¦-direction, we can write that the normal force π minus π times the cos of π minus πΉ times the sin of π is equal to the blockβs mass multiplied by its acceleration in the π¦-direction. Again, we know that that acceleration is zero, which means that π, the normal force magnitude, is equal to π cos π plus πΉ sin π.

If we substitute in for πΉ, π times the tangent of π, we now have an expression for π, what we want to solve for, in terms of known values π and π. When we plug in 1.50 kilonewtons for π and 30.0 degrees for π and calculate this expression, we find that π is 1.73 times 10 to the third newtons. Thatβs the magnitude of the normal force acting on the block.