A 4000N block is moved up a slope that is 20 degrees to the horizontal. If the coefficient of friction is 0.3, find the force parallel to the slope to: a) move the load up the slope b) prevent the load from sliding back

W=mg= 4000 or do i multiply this by 9.81

mg=4000 N is correct.

θ=20°
μ=0.3 (assumed static friction)

Gravity force along plane
F=mg sin(θ)

Friction force along inclined plane
=μmg cos(θ)
direction acts against movement.

To move load up slope:
Force required
=mg sin(θ)+μmg cos(θ)

To move load down slope:
Force required
=μmg cos(θ)-mg sin(θ)
if μmg cos(θ) > mg sin(θ)
=0 otherwise

F=4000sin(20)= 1368.1

FF= 0.3*4000cos(20)= 1127.6
FR= 4000sin(20) + 0.3*4000cos(20)= 2495.7
So 2495.7 N required to move it up the slope
1127.6 N is required to prevent the load from sliding back

Downward force due to gravity is

F=1368.1

Friction force resisting movement is
FF=1127.6

Therefore net downward force
=force required to prevent load from sliding down
=F-FF
=1368.1-1127.6
=240.45 N

To find the force parallel to the slope to move the load up and prevent it from sliding back, we need to consider the forces acting on the block.

Let's break down the forces acting on the block along the slope:

1. The force of gravity acting vertically downwards can be divided into two components: one perpendicular to the slope and one parallel to the slope.
- The component perpendicular to the slope is given by the equation: F_perpendicular = mg * cos(theta), where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The component parallel to the slope is given by the equation: F_parallel_gravity = mg * sin(theta).

2. The normal force (N) acts perpendicular to the slope. Its magnitude is equal to the perpendicular component of the weight: N = mg * cos(theta).

3. The force of friction (F_friction) opposes the motion of the block on the slope. Its magnitude is given by the equation: F_friction = coefficient of friction * N.

Now let's solve the two parts of the question:

a) Force parallel to the slope to move the load up the slope:
To move the load up the slope, the force parallel to the slope should overcome the force parallel to gravity. Therefore, the force parallel to the slope can be calculated by subtracting the force parallel to gravity from the force of friction:
F_parallel_move = F_friction - F_parallel_gravity
= (coefficient of friction * N) - (mg * sin(theta))

b) Force parallel to the slope to prevent the load from sliding back:
To prevent the load from sliding back, the force parallel to the slope should balance the force parallel to gravity, considering the opposite direction:
F_parallel_prevent_slide = F_parallel_gravity - F_friction
= (mg * sin(theta)) - (coefficient of friction * N)

Substituting the given values:
- Mass (m) is unknown.
- Weight, W = mg = 4000 N.
- Angle, theta = 20 degrees.
- Coefficient of friction, μ = 0.3.

We can assume g = 9.8 m/s^2.

First, we need to calculate the mass to proceed further. Using the equation W = mg, we can find the mass:
4000 N = m * 9.8 m/s^2
m = 4000 N / 9.8 m/s^2

After calculating the mass, we can substitute it into the equations to find the values of F_parallel_move and F_parallel_prevent_slide.