A cord holds stationary a block of mass m = 15 kg, on a friction less plane that is inclined at angle 6 = 27°.
.a) What are the magnitudes of the force f On the block from the cord and the normal force N on the block from the plane?

SOLUTION: Those two forces and the gravitational force FR on the block are shown in the block’s free body diagram . Only these three forces . A Key Idea is that we can relate them to the block’s acceleration via Newton’s second Law .We write as

With two unknown vector in , we cannot solve it for either vector directly on a vector-capable calculator. So. we must rewrite it in terms of components. We use a coordinate system with its x axis parallel to the plane, as shown in then two forces (N and f) line up with the axes.

from which,

(b) We now cut the cord. As the block then slides down the inclined plane, does it accelerate? If so, what is its acceleration?

SOLUTION: Cutting the cord removes force f from the block. Along the y axis, the normal force and component are still in equilibrium. However, along the x axis, only force component acts on the block; because it is directed down the plane (along the x axis), that component must cause the block to accelerate down the plane. Our Key Idea here is that we can relate to the acceleration a that it produces with Newton’s second law written for x components. We get

The magnitude of this acceleration a is less than the magnitude of the free fall acceleration because only a component of the component that is directed down the plane is producing acceleration a.

CHECKPOINT 7: In the figure, horizontal force F is applied to a block on a ramp. (a) Is the component of F that is perpendicular to the ramp F cos 8 or F sin ? (b) Does the presence of F increase or decrease the magnitude of the normal force on the block from the ramp?