Question

A 4.00-kg block rests on a 30.0° incline as shown in the figure. If the coefficient of static friction between the block and the incline is 1.70, what magnitude horizontal force F must act on the block to start it moving up the incline?

Answer 1

83.26 N

Step-by-step explanation:

Let F be the horizontal force applied to the block . This has two components:

`F_(\parallel)`
`\textrm {which will act upward parallel to the plane}` and

`F_(\perp) \textrm{ which is the forcce that will act downward on an inclined plane perpendicular to the plane. }`

`F_(\parallel) = F\cos\theta \\\\\\F_\perp} = F\sin\theta`

The weight of the block is mg acting vertically downward.
mg = 4 x 9.8 = 39.2 N

This can be decomposed into 2 components - one parallel to the plane acting downward opposite to
`F_(\parallel)` and the other perpendicular to the plane and acting in the same direction as
`F_(\perp)`

The two components of mg are

`mg\cos\theta = mg\cos30 = 39.2 (0.866) = 33.95 N`

`mg\sin\theta = mg\sin30 = 39.2 (0.5) = 19.6 N`

In addition we have the friction component that acts on the perpendicular combined force

Total perpendicular force =
`F\sin\theta + mg \cos\theta = F\sin30 + mg\cos30\\`

This perpendicular force also experiences friction with a coefficient of 0.7 so net perpendicular force =

`0.7(F\sin\theta + mg \cos\theta) = 0.7(F\sin30 + mg\cos30)\\\\`

Total Force acting parallel to the plane
=
`\rm F_(\parallel) - mg\sin\theta since mg\sin\theta \textrm{ is in a direction opposite to the parallel component of F}`

So we get

`\rm Fcos30 - mgsin30 = 0.7(Fsin30+5gcos30)\\`

Or

`\rm Fcos30 = mgsin30 + 0.7(Fsin30+mgcos30)\\`
Plugging in values for sin and cos θ and mg we get

F x 0.866 = 39.2 x 0.5 + 0.7(F x 0.5 + 39.2 x 0.866)

==> 0.866F = 19.6 + 0.35F + 23.76

==> 0.866F = 0.35F + 42.96

==> 0.866F - 0.35F = 42.96

==> F(0.866 - 0.35) = 42.96

==> 0.516F = 42.96

==> F = 42.96/0.516 =

==> F ≈ 83.26 N