Question

A box of mass 12 kg is dragged, with a constant acceleration of 1.75 m/s², up a path inclined at 30° to the horizontal. The force pulling the box has magnitude 2X N and acts at 10° to the path. The frictional force has magnitude X N. Calculate the value of X and the magnitude of the normal contact force of the path on the box.

Answer 1

The frictional force (X) is calculated to be approximately 15.3 N and the normal contact force on the box from the path is approximately 184.45 N by resolving forces and using Newton's second law.

Step-by-step explanation:

Calculating the Value of X and the Normal Contact Force on an Inclined Plane

To find the value of the frictional force (X) and the normal contact force, we can use Newton's second law and some trigonometry. The box is accelerated up an inclined plane, so the net force can be determined by the mass (12 kg) and the acceleration (1.75 m/s2). The pulling force is at an angle, which means we need to resolve it into parallel and perpendicular components to the incline.

First, let's find the parallel component of the pulling force (2X N) up the incline:

• 2X * cos(10°) - X - (mg sin(30°)) = mass * acceleration
• 2X * cos(10°) - X - (12 kg * 9.81 m/s2 * 0.5) = 12 kg * 1.75 m/s2

Solve for X, this will give us the frictional force:

• X ≈ 15.3 N (the force of friction)

Next, we can find the normal contact force exerted by the path:

• Normal force = 2X * sin(10°) + mg cos(30°)
• Normal force ≈ 184.45 N

The value of X is approximately 15.3 N and the normal contact force is approximately 184.45 N.