Calculating Force P: Solving a Statics Problem with Friction and Pulley Neglect

[smashbrohamme]

Knowing that friction equals 0.2 at all surfaces of contact, determine the magnitude of the force P required to move the 40kg plate B to the left. (Neglect Pulley Friction)

This is a statics problem asking how much force will it take to move block B which is 40kg...the problem is it has another block of 80kg sitting on top of it...and a pulley is connecting the 80kg block and the 40kg block.

the answer is 549N but I can't seem to find how to get it.


the 40kg block is on top of a friction plane that is 0.2...and a block is sitting on top of it that weighs 80kg. so I am assuming you have to count both frictions into the equation...

for the 40kg block I got Force to move it by itself is 152.96...and I am getting it takes 156.96 to move the 80kg block as well. so if you neglect the pulley's friction and just create a Tension of 156.96 I am only getting a force of 313.92N.

the picture is a lil funky but there is no angles to worry about.

 

[p21bass]

Draw a FBD for each block. You'll see that there are two friction planes for the bottom block, plus the tension in the string.

 

[smashbrohamme]

I believe I did account for both frictions of the 40 kg block, I am still not coming up with the correct answer that was given. would the friction from both top and bottom of the 40kg block be going opposite of the p(force)

 

[PhanthomJay]

I believe I did account for both frictions of the 40 kg block, I am still not coming up with the correct answer that was given. would the friction from both top and bottom of the 40kg block be going opposite of the p(force)

Yes, the friction always opposes the direction of relative motion or impending relative motion between the 2 objects or surfaces in contact with the object you are examining for forces. Follow p21bass' advice, and be sure you are calculating the normal force between the lower block and floor correctly.

 

[rwisz]

I would first clarify the problem statement and make sure that all the given information is accurate and complete. From the given information, it seems that we are dealing with a system of two blocks connected by a pulley, with the 40kg block on top of the 80kg block. The friction coefficient between the 40kg block and the surface is 0.2, and the pulley friction is neglected.

To calculate the force required to move the 40kg block to the left, we need to consider the forces acting on both blocks. The 80kg block exerts a downward force due to its weight, which is equal to 80kg x 9.8m/s^2 = 784N. This force is balanced by the normal force from the surface, which is also equal to 784N.

Now, let's consider the 40kg block. The only force acting on it is the tension from the pulley, which is equal to the force required to move the 80kg block, i.e. 784N. However, we also need to take into account the friction force acting on the 40kg block. The maximum friction force is given by the product of the coefficient of friction and the normal force, which in this case is 0.2 x 784N = 156.8N.

To move the 40kg block, we need to overcome this friction force, in addition to the force required to move the 80kg block. Therefore, the total force required is 784N + 156.8N = 940.8N. However, this is the force required to move the system as a whole, so we need to divide by the mechanical advantage of the pulley system, which is 2. This gives us a final answer of 940.8N / 2 = 470.4N.

It is important to note that this force is the minimum required to move the system. In reality, we would need to apply a slightly higher force to overcome the initial static friction and get the system moving. This value would depend on the specific properties of the surfaces in contact and cannot be accurately calculated without more information. Therefore, the answer of 549N given in the problem statement may be an approximation or an experimental result.

In conclusion, to accurately calculate the force required to move the 40kg block to the left, we