Question

A box at rest is in a state of equilibrium half way up on a ramp. the ramp has an incline of 42° . what is the force of static friction acting on the box if box has a gravitational force of 112.1 n ?70 n80 n75 n85 n

Answer 1

The force of static friction acting on the box can be calculated using the component of gravitational force parallel to the incline, which balances the static friction when the box is in equilibrium.

Step-by-step explanation:

The question presented is a physics problem concerning static friction and a box in equilibrium on an inclined plane. Since the box is at rest and in equilibrium, the force of static friction must exactly balance the component of gravitational force acting parallel to the incline. This component can be calculated using the formula F_gravity_parallel = F_gravity * sin(\theta), where \theta is the angle of incline and F_gravity is the gravitational force acting on the box. As the gravitational force provided is 112.1 N and the incline of the ramp is 42°, we can calculate this component as:

F_gravity_parallel = 112.1 N * sin(42°)

This yields a component of gravitational force parallel to the incline that will be equal to the force of static friction. To find the correct answer from the provided options, one would calculate this using a calculator and choose the option closest to the calculated value.

Answer 2

The force of static friction acting on the box is approximately
`\(75 \, \text{N}\).`

To find the force of static friction acting on the box, we'll need to consider the forces acting along the incline of the ramp.

First, let's determine the component of the gravitational force parallel to the ramp's surface. This force, known as the component of the weight force along the incline
`(\(F_{\text{parallel}}\))`, can be calculated using trigonometry:

`\[ F_{\text{parallel}} = F_{\text{gravity}} * \sin(\text{angle of incline}) \]`

Given that the gravitational force
`\(F_{\text{gravity}}\)` is 112.1 N and the angle of incline is 42°:

`\[ F_{\text{parallel}} = 112.1 * \sin(42^\circ) \]`

Let's calculate
`\(F_{\text{parallel}}\):`

`\[ F_{\text{parallel}} \approx 112.1 * 0.669 \approx 75 \, \text{N} \]`

Now, since the box is at rest and in equilibrium halfway up the ramp, the force of static friction
`(\(F_{\text{friction}}\))` must be equal in magnitude and opposite in direction to
`\(F_{\text{parallel}}\)` to prevent the box from sliding down or up the ramp.