The magnitude of the minimum force applied perpendicularly to the incline to prevent the crate from sliding?

Question

crate - 3.00kg
incline- 35.0 degree above the horizontally axis
coefficient of static friction - 0.300

Answer 1

Okay, I drew the picture. The mass of the crate is 3.00 kg and its vertical force W = 3.00 g = 3.00 x 9.8 = 29.4 N.

Resolve that into normal and parallel components with respect to the inclined plane.

The normal (perpendicular) component is

Wn = W cos 35

and the (downward) parallel component is

Wp = W sin 35

It's the downward parallel component that wants to push the crate down the plane. This must be offset by an equal and opposite parallel frictional force pushing up the plane.

Let F be the external normal force applied to the crate. Then the total normal force applied to the crate is

F + Wn = F + W cos 35

The coefficient of static friction is 0.300, so the total frictional force is

0.300(F + W cos 35)

and for equilibrium, we have

0.300(F + W cos 35) = W sin 35

Solving for F, we have

(3/10)(F + W cos 35) = W sin 35
F + W cos 35 = (10W/3) sin 35
F = W [(10/3) sin 35 - cos 35]
F = 32.127421 N

which you can round off.

Answer 2

The component of the weight of the crate acting down the incline = m*g*sin theta = 3*9.8*sin 35 = 16.86 N
This should be compensated by the frictional force, which is = coeff of friction * normal reaction.
Hence normal reaction = 16.86/.3 = 56.21 N
Out of this, the normal component of the weight of the crate accounts for m*g*cos theta = 24.08 N
The balance is the normal force that is required = 32.13 N

Answer 3

Corrections to the respond given by Pearlsaw... the traditional rigidity N = F + mgcos25, the rigidity of friction: F(f) = (mu)(F + mgcos25). be conscious: incorrect placement of parenthesis, the utilized rigidity F is perpendicular to the incline, no longer down like the rigidity of gravity. fixing for F we've : F = 15.9 N

Answer 4

The sliding force Fs = g*m*sin(35deg) = 16.875 N. The gravitational friction force Ff = g*m*cos(35deg)*0.3 = 7.230 N. Thus another normal force must be added such that Ff>=Fs. The added normal force required is Fa>=(Fs-Ff)/0.3, or
Fa >= sin(35deg) - g*m*(0.3*cos(35deg)) / 0.3 = 32.1492 N.
Note I'm using g=9.80665 m/s^2, the accurate value.