A block with a mass of 3.07 kg is placed at rest on a surface inclined at an angle of 44.6° above the horizontal. The coefficient of static friction between the block and the surface is 0.540, and a force of magnitude pushes upward on the block, parallel to the inclined surface. The block will remain at rest only if is greater than a minimum value, Fmin, and less than a maximum value, Fmax. Calculate Fmin.

Question

A block with a mass of 3.07 kg is placed at rest on a surface inclined at an angle of 44.6° above the horizontal. The coefficient of static friction between the block and the surface is 0.540, and a force of magnitude pushes upward on the block, parallel to the inclined surface. The block will remain at rest only if is greater than a minimum value, Fmin, and less than a maximum value, Fmax. Calculate Fmin.

Answer 1

To keep the block from sliding downhill the block uphill, the applied force must exceed

Fmin = M*g*sin44.6 - M*g*cos44.6 * 0.540

That is the component of weight down the hill, minus the maximum possible friction force

Answer 2

To calculate the minimum force, Fmin, needed to keep the block at rest, we need to consider the forces acting on it.

First, let's analyze the forces in the vertical direction. Since the block is at rest and not moving vertically, the vertical forces must be balanced.

The force acting downward due to gravity can be calculated using the formula:
F_g = m * g
Where:
m = mass of the block = 3.07 kg
g = acceleration due to gravity = 9.8 m/s^2

Substituting the values:
F_g = 3.07 kg * 9.8 m/s^2 = 29.986 N

Next, let's consider the forces acting parallel to the inclined surface. The only force in this direction is the frictional force.

The maximum static frictional force can be found using the formula:
F_friction_max = µ_s * F_N
Where:
µ_s = coefficient of static friction = 0.540 (given)
F_N = normal force

The normal force can be calculated as:
F_N = F_g * cos(theta)
Where:
theta = angle of inclination = 44.6°

Substituting the values:
F_N = 29.986 N * cos(44.6°) = 21.217 N

Now we can calculate the maximum static frictional force:
F_friction_max = 0.540 * 21.217 N = 11.461 N

Since the block is at rest on the inclined surface, the force pushing upward must be equal to or greater than the maximum static frictional force to prevent the block from sliding down.

Therefore, the minimum force, Fmin, is equal to the maximum static frictional force:
Fmin = 11.461 N

So, the minimum force required to keep the block at rest is 11.461 N.