p05_043.pdf

43. The free-body diagram for each block is shown below. T is the tension in the cord and θ =30 ◦ is the

angle of the incline. For block 1, we take the + x direction to be upthe incline and the + y direction

to be in the direction of the normal force N that the plane exerts on the block. For block 2, we take

the + y direction to be down. In this way, the accelerations of the two blocks can be represented by the

same symbol a , without ambiguity. Applying Newton’s second law to the x and y axes for block 1 and

to the y axis of block 2, we obtain

−

m 1 g sin θ = m 1 a

−

m 1 g cos θ =0

m 2 g

−

T = m 2 a

respectively. The ﬁrst and third of these equations provide a simultaneous set for obtaining values of a

and T . The second equation is not needed in this problem, since the normal force is neither asked for

nor is it needed as part of some further computation (such as can occur in formulas for friction).

(+ x )

•

m 2 g

(+ y )

m 1 g

(a) We add the ﬁrst and third equations above: m 2 g

−

m 1 g sin θ = m 1 a + m 2 a . Consequently, we ﬁnd

a = ( m 2 −

m 1 sin θ ) g

m 1 + m 2

= (2 . 30kg)

3 . 70sin30 . 0 ◦ )(9 . 8)

3 . 70 + 2 . 30

=0 . 735 m / s 2

(b) The result for a is positive, indicating that the acceleration of block 1 is indeed up the incline and

that the acceleration of block 2 is vertically down.

T = m 1 a + m 1 g sin θ =(3 . 70)(0 . 735) + (3 . 70)(9 . 8)sin30 ◦ =20 . 8N .

−