Question

Example #2A block with mass m = 5 kg sits on a surface with a coefficient of static friction sk= 0.5 and a coefficient of kinetic friction uk= 0.3. a)If you can pull on the block at any angle, what minimum force is required to break static friction and cause the block to slide? b)What is the optimal angle to pull at? c)If you pull at the optimal angle with the minimum force, what will the acceleration of the block be once static friction is broken?

Answer 1

along vertical

Fn + F*sintheta = mg

normal force , Fn = m*g - F*sintheta

frictional force fs = us*Fn = us*(m*g - F*sintheta)

when the friction breaks

fs = F*costheta

us*mg - us*F*sintheta = F*costheta

F*(us*sintheta + costheta) = us*m*g

F = us*m*g / (us*sintheta + costheta)

for F to be minimum

dF/dtheta = 0

us*costheta - sintheta = 0

tantheta = us

theta = tan^-1(us)

theta = 26.56 degrees

when the force is applied at optimal angle theta = 26.56

along vertical

Fn + F*sintheta = mg

normal force , Fn = m*g - F*sintheta

frictional force fs = us*Fn = us*(m*g - F*sintheta)

when the friction breaks

fs = F*costheta

us*(m*g - F*sintheta) = F*costheta

0.5*((5*9.8) - (F*sin26.56)) = F*cos26.56

minimum force F = 21.9 N   <<<----------------answer

optimal angle theta = 26.56 degrees

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(c)

Fnet = m*a

F*costheta - fk = m*a

F*costheta - uk*Fn = m*a

F*costheta - uk*(m*g - F*sintheta) = m*a

21.9*cos26.56 - 0.3*((5*9.8) - 21.9*sin26.56) = 5*a

acceleration a = 1.565 m/s^2