A mass M in held in place by the applied force F and the pulley system shown below. The pulleys are massless and frictionless. Determine the tension in each section in each section of rope and the applied force F
Taking upwards as positve
-T1 - F = 0
T4 - T1 - T3 = 0
T2 + T3 - T5 = 0...
In Figure 6-63, a block weighing 22 N is held at rest against a vertical wall by a horizontal force of magnitude 60 N. The coefficient of static friction between the wall and the block is 0.55, and the coefficient of kinetic friction between them is 0.38. In six experiments, a second force is...
I do not understand this problem, and since this is my first 3D problem, I can't figure out how to draw a FBD either, so I can't figure out how many forces acting on the crate. It'd be nice if someone could help me out. Thanks
So, I have created some blocks, and draw the FBD myself, am not sure if I did it right or not, so I'm going to upload these FBD here, and hope you guys can correct me if I'm wrong. Thanks a lot.
A block of m1 and m2, m2 is siting on top of m1. Assume there is no friction on the ground under m1...
Yes, now I got it, so basically if it's put on the ground, 12N is the minimum force that you need to apply in order to move it. So fs maximum will be 12 N, right ? and since we have fs = us * mg = 12N, we could find coefficient of static friction....uhm, or coefficient of kinetic friction ...
A block of mass m1 = 4 kg is put on top of a block of mass m2 = 5 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless...
Alright, so for m1, it should be fk = m1a ;)
Very clear explanation, wish that you were our professor college, physics would be fun then ;) Thank you very much for your help.
Thanks collinsmark, I finally got it and understand how third newton's law works. On the other hand, I'm still a bit confused about the sign of the frictional force acting on m1.
I uploaded the Picture and my FBD, and as you can see, if m2 is moved from right to left, m1 will be sliding from...
I think it will be opposite side with the horizontal force F ?
so fk2 = -m1a
a = -fk2 / m1 = -uk(m2g) / m1 ???
I still don't get it why frictional force fk2 will be the pulling force on m1, weird...
I'm sorry to bring it up again, but I'm having the same problem and reading this post helps me figure out how to do it; however, I think that what you're saying is not right Doc Al.
I draw a FBD of m and M, and here are all the forces that I've found:
Taking downwards as positive, x+ from left...
A slab of mass m1 = 40 kg rests on a frictionless floor. a block of mass m2 = 10 kg rests on top of the slab ? Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. The block is pulled by a horizontal force with a magnitude of 100 N...
Oh I got it, since acceleration is not constant, we can't use all the kinematic equations to solve this problem.....But how are we gonna find acceleration if we only have one equation ?
A 1000kg boat is traveling at 90 km/h when its engine is shut off. The magnitude of the frictional force fk between boat and water is proportional to the speed V of the boat: fk = 70V where V is in meters per second and fk is in newtons. Find the time required for the boat to slow to 45 km/h...
5. A 12N horizontal force F vector pushes a block weighing 5N against a vertical wall. The coefficient of static friction between the wall and the block is .60 and the coefficient of kinetic friction is .40. Assume the block is not moving initially.
(A) will the block start moving?
(B) What...
Ok, so static friction fs can vary, depending on the applied force and kinetic friction fk is constant as long as the block is in motion, and is independent with the applied force ? Is that right ?
Ok, so we use θ instead of a
we should have:
-P + fk - (mg)sinθ = ma
if you call F is the net force acting of the mass, we will have
F = -P + fk - (mg)sinθ = ma
The problem asks for the frictional force on the block from the plane which is fk, however, without knowing acceleration a...
Ok, so I have the same problem with 1MileCrash, and I don't wanna start a new thread, so I'm going to show you how I did:
a: angle of the inclined plane
y direction:
(-mg)cosa + N = 0
N = (mg)cosa
x direction:
-P + fs - (mg)sina = 0 (1)
P = fs - (mg)sina
P = us(N) - (mg) sina
P = 10.08N
So If...
I have the same problem too, and I don't really understand it. This is how I approached the problem. I first took into account the mass of the Motorcycle and the mass of the rider as well. So I tried to do FBD on each of them
Motorcycle
-m1gsin 10 + F = m1a
Rider
-mgsin10 = ma
Now I...
Oh, yeah,
Both Fa and Fb have the same direction, going from left to right.
So it should be
Tension in the string
4.0 kg block A and 6.0kg block B are connected by a string of negligible mass. Force a Fa = 12N acts on block A; force B Fb = 24N acts on block B. What is the tension in the...
4.0 kg block A and 6.0kg block B are connected by a string of negligible mass. Force a Fa = 12N acts on block A; force B Fb = 24N acts on block B. What is the tension in the string.
I was able to draw FBD and lists all the forces acting on A and B, but what I don't understand is the tension T...
two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0, container 1 has mass 1.30 kg and container 2 has mass 2.8 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.200 kg/s.
a) At what...
Sorry if I bring it up, but I have the same problem, and I'm trying to figure out how to solve it. I read all the post above, and one thing I don't understand is, the acceleration in those 2 figures should be different, they can't be the same, and if they're not the same, how can we set up the...