A block weighing 70.7 N rests on a plane inclined at 21.6° to the horizontal. The coefficient of the static and kinetic frictions are 0.23 and 0.12 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?
I STILL can't get this one. I've tried multiple answers, none of which are correct. Could someone please give me a little help?
When you are considering the force necessary to maintain a static condition, you use the static coefficient of friction. The maximum possible friction force is in that case
70.7*cos21.6 *0.23 = 15.12 N, The component of the gravity force down the plane is M g sin 23.6 = 70.7*0.400= 28.31 N.
Since the gravity component down the plane exceeds the maximum friction force by 13.19 N, that is the minimum force need to keep it from sliding.
Ok I understand that part.
Now it asks for the minimum magnitude of F that will start the block moving up the plane, and the magnitude of F that is required to move the block up the plane at constant velocity. Is it a similar process for both?
When you talk about a moving situation, the kinetic friction coefficient applies. The friction force is down the plane when trying to pull the block up the plane, so
70.7 sin 23.5 + 70.7 cos 23.5 * 0.12 = F
(For constant velocity, the forces balance)
Solve for F
Do you come up with an answer of 33.91? Because that's what I get.
No, I get 35.97 N
Why is it sin and cos of 23.5? I thought the angle was 21.6...
Yes. My mistake
Using 21.6 as the angle for both cos and sin, I come up with 33.91, which is the incorrect answer. Am I still doing something wrong?
Certainly! Let's break down the problem step by step.
First, let's calculate the normal force (N) acting on the block. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, it will be the component of the weight of the block perpendicular to the inclined plane.
The weight of the block (W) can be calculated using the formula W = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the weight of the block is 70.7 N, we can calculate the mass (m) of the block by rearranging the equation as m = W/g.
m = 70.7 N / 9.8 m/s² ≈ 7.2 kg
Now, we can calculate the normal force (N) using the formula N = mg. Plug in the values:
N = 7.2 kg × 9.8 m/s² ≈ 70.6 N
Next, let's find the force of friction (Ff) acting on the block. The force of friction can be calculated using the equation Ff = μN, where μ is the coefficient of friction.
Given that the coefficient of static friction (μs) is 0.23, we can calculate the static frictional force (Ffs) using the formula Ffs = μsN.
Ffs = 0.23 × 70.6 N ≈ 16.2 N
Since the block is at rest, it experiences static friction opposing the motion. The static frictional force can take any value from zero up to its maximum value. In this case, the maximum static frictional force (FfsMax) is equal to Ffs.
FfsMax = Ffs = 16.2 N
Finally, we need to find the minimum magnitude of the force (F) parallel to the plane that will prevent the block from slipping. To do this, we equate the force of friction (Ff) to the applied force (F).
Ff = F
Since we are looking for the minimum value of F, we substitute FfsMax for Ff.
FfsMax = F
Therefore, the minimum magnitude of the force F, parallel to the plane, is 16.2 N.
I hope this helps! Let me know if you have any further questions.