# Force and motion

1. A 8 kg block of steel is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.45. A force is to be applied to the block. To three significant figures, what is the magnitude that applied force if it puts the block on the verge of sliding when the force is directed a) upward at 60 degree from the horizontal

b) downward at 60 degree from the horizontal

2. A 11kg block of steel is at rest on a horizontal table. The coefficient of static friction between block and table is 0.52

a) What is the magnitude of a force acting upward 60 degree from the horizontal that will put the block on the verge of moving? (answer: 59N) Please explain.

b) if the force acts downward at 60 degree from the horizontal, how large can its magnitude be without causing the block to move? (answer: $1.1*10^3N$) Please explain.

1. The horizontal component of the applied force must equal the friction force on the block. In both a and b cases the applied force must be decomposed on the x and y (horizontal and vertical) axes.

a) $F y = F*sin(\alpha)$ , $F x = F*cos(\alpha)$

$F x =\mu(m*g -F y)$

$F*cos(\alpha) =0.45(8*9.81 -F*sin(\alpha))$

$F(cos(60)+0.45*sin(60)) = 0.45*8*9.81$

$F = 39.694 N$

b)

$F y = F*sin(\alpha)$, $F x =F*cos(\alpha)$

$F x = \mu(m*g +F y)$

$F*cos(\alpha) = 0.45(8*9.81 +F*sin(\alpha))$

$F(cos(60) -0.45*sin(60)) =0.45*8*9.81$

F = 320.215 N

2. The horizontal component of the applied force must equal the friction force on the block. In both a and b cases the applied force must be decomposed on the x and y (horizontal and vertical) axes.

a) $F y = F*sin(\alpha)$ , $F x = F*cos(\alpha)$

$F x =\mu(m*g -F y)$

$F*cos(\alpha) =0.52(11*9.81 -F*sin(\alpha))$

$F(cos(60)+0.52*sin(60)) = 0.52*11*9.81$

$F = 59.046 N$

b)

$F y = F*sin(\alpha)$, $F x =F*cos(\alpha)$

$F x = \mu(m*g +F y)$

$F*cos(\alpha) = 0.52(11*9.81 +F*sin(\alpha))$

$F(cos(60) -0.52*sin(60)) =0.52*11*9.81$

F = 1129.793 N