Question

A mass of 8 kg rests on a floor with a coefficient of static friction of 0.88 and coefficient kinetic friction of 0.33. One force is applied vertically downward of 28 Newtons and another force is applied at an angle of 26 degrees below the horizontal. Calculate what the magnitude this force needs to be in order to begin to move the mass. If that force is removed and another force is applied at an angle of 26 degrees above the horizontal (the downward force is still there), calculate the force needed to begin to move the force for that situation. How much more force is needed to begin the mass, in Newtons, when the force is angled below the horizontal compared to above the horizontal?

Answer 1

Given mass =8kg

= 0.88 and   = 0.33

Vertical force = 28N

Case 1). Force F acting at an angle 26 deg BELOW the horizontal

Lets split force F into its components

Fx = Fcos 26 acting in +ve x direction

Fy = Fsin26 acting downwards in -ve y direction

When the mass starts to move Force F will be able to overcome frictional force.

Fnet =Fx -F_friction = 0

Consider only (there is no movement)

Fx=F_friction --------------(1)

F_friction =   *Normal reaction

N= mg+ vertical force +Fy(acting downwards)

= 8*9.81 + 28 + Fsin26

From equ (1)

Fcos26 = 78.48 + 28 +Fsin26

F(cos26-sin26)= 106.48

F = 106.48/0.46 =231.478 N

ANSWER F = 231.478 N

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Case 2) Force F acting at an angle 26 deg ABOVE the horizontal

Fx = Fcos 26 acting in +ve x direction

Fy = Fsin26 acting upwards in +ve y direction

When the mass starts to move Force F will be able to overcome frictional force.

Fnet =Fx -F_friction = 0

Fx=F_friction --------------(2)

F_friction =   *Normal reaction

N= mg+ vertical force -Fy(acting upwards)

= 8*9.81 + 28 - Fsin26

From equ (1)

Fcos26 = 78.48 + 28 -Fsin26

F(Cos26 +sin 26) = 106.48

F = 106.48/1.337 =79.64 N

ANSWER : F = 79.64N

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ANSWER last part

231.478 - 79.64 =151.838 N

We require an extra 151.838 N in the first case where force is angled BELOW the horizontal when compared to force F angled ABOVE the horizontal