A 2.00 kg block, on a horizontal surface with coefficient of static friction 0.900, is pushed by a force that is directed down at an angle of 30.0 degrees
to the horizontal. Find the minimum magnitude of the force needed to get the block to start moving.
so I found my N force it's 19.6N and I calculated fsmax to be 17.64N, but I'm not sure what to do now.
oh I think I have found the answer! Sorry
To find the minimum magnitude of the force needed to get the block to start moving, you need to compare the maximum static friction force to the force applied at an angle.
1. Start by calculating the vertical component of the force applied, which is the force multiplied by the sine of the angle:
F_vertical = F * sin(30°)
2. Calculate the horizontal component of the force applied, which is the force multiplied by the cosine of the angle:
F_horizontal = F * cos(30°)
3. Determine the maximum static friction force, which is the product of the coefficient of static friction and the normal force:
fs_max = μ * N
With μ = 0.900 and N = 19.6 N, you correctly calculated fs_max to be 17.64 N.
4. Now, compare the horizontal component of the force applied to the maximum static friction force:
F_horizontal ≥ fs_max
Substituting the values, we get:
F * cos(30°) ≥ 17.64 N
Rearrange the equation to solve for F:
F ≥ 17.64 N / cos(30°)
5. Finally, substitute the value of cos(30°) ≈ 0.866 and calculate the minimum magnitude of force needed:
F ≥ 17.64 N / 0.866
Therefore, the minimum magnitude of the force needed to get the block to start moving is approximately 20.37 N.