A Classical Approach of Newtonian Mechanics

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4 NEWTON’S LAWS OF MOTION 4.9 Frames of reference


F

m

25 o

Combining the above two expressions, making use of the fact that T 4 = M g, gives


M =

T 2
.
g tan 40 ◦

Finally, since T 2 = 50 N and g = 9.81 m/s^2 , we obtain
50
M =
9.81 × 0.8391


= 6.074 kg.

T 1

T 2

T 4

Worked example 4.2: Block accelerating up a slope


Question: Consider the diagram. Suppose that the block, mass m = 5 kg, is


subject to a horizontal force F = 27 N. What is the acceleration of the block up
the (frictionless) slope?


Answer: Only that component of the applied force which is parallel to the incline


has any influence on the block’s motion: the normal component of the applied
force is canceled out by the normal reaction of the incline. The component of


the applied force acting up the incline is F cos 25 ◦. Likewise, the component of


40 o
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