What push parallel to a 42 incline is required to just start a 340 N block sliding up the plane? us = 0.6

x: 0 =F-m•g•sinα-F(fr)

y: 0=N-m•g•cosα,

0 =F-m•g•sinα-F(fr)= F-m•g•sinα- μ•N=
=F - m•g•sinα- μ•m•g•cosα.
F = m•g(sinα- μ•cosα)=
= 340•(sin42-0.6•cos42)=...

To find the push parallel force required to just start a block sliding up the incline, we need to consider the forces acting on the block.

1. Weight: The weight (W) of the block acting vertically downwards is given by:
W = mass * gravity
In this case, the weight is equal to the force of gravity acting on the block, which is W = 340 N.

2. Normal force: The normal force (N) is the force exerted by the incline on the block perpendicular to the frictional force. Since the block is on an incline, the normal force is not equal to the weight. It can be calculated as:
N = W * cos(θ)
Here, θ is the angle of inclination, which is 42° in this case.

3. Frictional force: The frictional force (f) acts parallel to the incline and opposes the block from moving. The magnitude of the frictional force is given by:
f = u_s * N
Here, u_s is the coefficient of static friction, which is given as 0.6.

Now, to find the push parallel force that just starts the block sliding up the incline, we need to overcome the static frictional force (fs). When the force applied is just enough to overcome the static friction, the block starts moving.

Thus, the push parallel force required is equal to the maximum static frictional force (fs) acting on the block. Mathematically, fs can be calculated as:
fs = u_s * N

Substituting the values we have:
fs = 0.6 * W * cos(θ)

Let's calculate the push parallel force:
fs = 0.6 * 340 * cos(42°)

fs ≈ 0.6 * 340 * 0.7431
fs ≈ 148.164 N

Therefore, the push parallel force required to just start the block sliding up the incline is approximately 148.164 N.