Question

A horizontal force F~ is applied to a block of mass m = 1 kg placed on an inclined

Answer 1

Hi there!

To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.

Recall:

$\large\boxed{F_D = \mu N}}$

The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.

$\large\boxed{N = Mgcos\theta + Fsin\theta}}$

We can plug in the known values to solve for one part of the normal force:

N = (1)(9.8)(cos30)  + F(.5) = 8.49  + .5F

Now, we can plug this into the equation for the dynamic friction force:

Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F

For a block to move with constant speed, the summation of forces must be equivalent to 0 N.

If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:

Fcosθ = 1.697 + .1F

Solve for F:

Fcos(30) - .1F = 1.697

F(cos(30) - .1) = 1.697

F = 2.216 N