# Transcript: Calculating the ramp angle quiz

**Diagram:**

A right triangle is formed by the ground, the stage, and the ramp, with the ramp as the hypotenuse sitting at an unknown angle to the ground, represented by the theta symbol. Sitting on the hypotenuse of the triangle is the trolley. A vector pointing downward from the front of the trolley, parallel to the hypotenuse, represents the force pulling the trolley down the ramp, which is 16 kilograms. A vector extending upward from the trolley perpendicular to the hypotenuse represents [latex]F_N[/latex], the normal force. A vector extending downward from the trolley perpendicular to the ground represents the weight force, expressed as [latex]m\times g[/latex]. The weight force and the force of the trolley on the ramp create another right triangle. The angle at which these vectors extend from the trolley is again unknown and represented by the theta symbol, and the second acute angle of the triangle is [latex]90^\circ - \theta[/latex].

## Question

If the total mass to be pulled up the ramp is 54kg, and the pulling force is 16kg, calculate the angle of the ramp. Assume friction on the ramp is negligible. Round your answer to the nearest degree.

θ =

## Hint

We can apply the same formula in question 1 to calculate the angle of the ramp. First, we need to convert the 16kg of force to newtons by multiplying by the gravity constant, [latex]9.8m/s^2[/latex]:

[latex]16kg \times 9.8m/s^2 = 156.8N[/latex]

[latex]\sin \theta = 156.8N \div mg[/latex]

[latex]\sin \theta = 156.8N \div 54kg \times 9.8m/s^2[/latex]

[latex]\sin \theta = 28.456[/latex] (note that all unites cancel out)

[latex]\sin^{-1} \theta = 17.2 \text { degrees}[/latex]

## Answer

θ = 27kg