A Classical Approach of Newtonian Mechanics

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