Final answer:
The gravitational force exerted on Earth by the block and the total force exerted by the incline on the block (if it includes both normal force and friction) have magnitudes equal to mg. However, the normal force on an incline is adjusted for the angle and is typically less than mg, as it counteracts only the gravitational component perpendicular to the incline. The frictional force is also typically less than mg.
Step-by-step explanation:
In the scenario where a block of mass m is at rest on a rough incline, the forces that must have a magnitude equal to mg, which is the weight of the block, are:
- The gravitational force exerted on Earth by the block (d).
- The total force exerted by the incline on the block, if it includes both the normal force and the force of friction balanced such that the block remains at rest (a), assuming the question implies equilibrium in the vertical direction.
However, please note that the normal force (b) alone does not always have a magnitude equal to mg on an incline. It needs to be adjusted for the angle of the incline since only a component of the gravitational force is perpendicular to the inclined surface. Therefore, in most cases, the normal force is less than mg because it only counteracts the component of gravity acting perpendicular to the incline.
The force of friction (c) is typically less than mg as it only needs to balance the component of gravity that acts parallel to the incline and is responsible for attempting to slide the block down the incline.