p05_072.pdf

72. Wetake+ x uphillforthe m =1 . 0kgboxand+ x rightwardforthe M =3 . 0kgbox(sotheaccelerations

ofthetwoboxeshavethesamemagnitudeandthesamesign). Theuphillforceon m is F andthedownhill

forces on it are T and mg sin θ ,where θ =37 ◦ . The only horizontal force on M is the rightward-pointed

tension. Applying Newton’s second law to each box, we ﬁnd

−

mg sin θ = ma

T = Ma

which are added to obtain F

−

mg sin θ =( m + M ) a . This yields the acceleration

a = 12

−

(1 . 0)(9 . 8)sin37 ◦

1 . 0+3 . 0

=1 . 53 m / s 2 .

Thus, the tension is T = Ma =(3 . 0)(1 . 53)=4 . 6N.