Question

A block of mass 5 kg is at rest on a ramp which is at 25◦. The coefficient of static friction between the block and the slope is 0.4. What is the minimum magnitude of the horizontal force, F, that is required to make the block move? Assume `g=10ms^-2`

Answer 1

`F = 53.2 N`

Explanation:

I now understand that downhill on the slope is to the left and the force is uphill -- to the right. So the goal is to move the block uphill. I will consider positive motion and forces to be uphill.

The friction between the block and slope is

`F_f = mu*N`

`N` is the normal force pressing the 2 surfaces together. The weight of the block and the applied force will contribute to N. N will be the total of the components, perpendicular to the slope, of the block's weight and of the applied force. So

`N = 5 kg*10 m/s^2*cos25^@ + F*sin25^@ = 45.3 N + F*sin25^@`

We need to understand that if we are trying to push the block up the slope, friction will fight that and therefore needs to have a negative sign. So `F_f` is

`F_f = -mu*(45.3 N + F*sin25^@)`

`F_f = -0.4*(45.3 N + F*sin25^@)`

The component of the block's weight that points down the slope is

`W_"ds" = m*g*sin25^@ = 5 kg*(-10 m/s^2)*sin25^@ = -21.1 N`

So `F_f and W_"ds"` are fighting against motion uphill. The total of the opposition is

`"opposition" = -0.4*(45.3 N + F*sin25^@) -21.1 N`

Now, about the force that is applied. The upslope component of that, `F_"us"` is

`F_"us" = F*cos25^@`

At the minimum value of the applied force, the sum

`F*cos25^@ - 0.4*(45.3 N + F*sin25^@) -21.1 N = 0`.

Solving for `F`

`F*cos25^@ - 18.1 N - 0.4*F*sin25^@ -21.1 N = 0`

`0.906*F - 18.1 N - 0.169*F -21.1 N = 0`

`0.737*F = 39.2 N`

`F = (39.2 N)/(0.737) = 53.2 N`

I hope this helps,
Steve