A block of mass 3kg is placed on a rough surface. The coefficient of static friction between two surfaces is 0.2 then what is the minimum horizontal force required to move the block?
As stated, there is not sufficient information to answer.What is the acceleration due to gravity whereas the experiment is performed?If you can assume 1 Earth standard gravity, the the block is pressed down onto the surface by a force of 9.8 * 3 N. The coefficient of friction presents a retarding force of 0.2 * 29.4, or 5.88N
The total force need to drag a box at constant speed across a surface with coefficient of kinetic friction mk is least...?
The total force need to drag a box at constant speed across a surface with coefficient of kinetic friction mk is least when the force is applied at an angle (theta) such that: a.) sin(theta) = mk b.) cos(theta) = mk c.) tan(theta) = mk d.) cot(theta) = mk e.) sec(theta) = mk
Box of books being dragged at speed of 1.5 m/s with coefficient of friction problem?
A box of books that weighs 40N is dragged at a speed of 1.5 m/s across a rough floor. If the coefficient of friction between the floor and the box is 0.20, what is the rate at which heat energy is dissipated?
A 50-kg box is pushed along a horizontal surface. The coefficient of kinetic friction is 0.35. What horizontal force would accelerate it at 1.2 m/s^2?
The situation involves taking into account two main contributing forces. These are the Friction Forces and the Accelerating forces.1. If the box has to be moved with constant velocity then only the frictional forces will be existing. ( 50. 9.8. 0.35) this assumes that the frictional forces remain constant and that the force is horizontal. If the force was at a down or up angle pushing or at an up or down angle pulling then the problem would add other components due to the up angle or down angle of the pushing force.2. When accelerating in addition to the friction forces one also needs to add an accelerating force which is M.a or (50. 1,2)Again it would add a little more effort if one considers the pushing or the pulling forces, not only when horizontally applied but at an angle. This would give an idea of what to select in actual real life when one is faced with this dilemma.
Find the magnitude of the horizontal force necessary to drag a block to the left at a constant speed of 8.00?
Block A weighs 1.80 and block B weighs 4.40 . The coefficient of kinetic friction between all surfaces is 0.300. (Block A is sitting on top of block B with the force of kinetic friction going to the left) Part A. Find the magnitude of the horizontal force necessary to drag block B to the left at a constant speed of 8.00 cm/s if A rests on B and moves with it. F(a)= ? N Part B. Find the magnitude of the horizontal force necessary to drag block B to the left at a constant speed of 8.00 cm/s if A is held at rest by a string (same as before, just A on a string to the right on a wall). F(b)= ? N Part C. In part (A), what is the friction force on block A? F(fric)= ? N ANY HELP WILL MAKE A DIFFERENCE!!!
A horizontal force of 150 N is used to push a 48.0 kg packing crate a distance of 4.85 m on a rough horizontal?
A) Remember that Work = Force * Distance So, W = 150(4.85) = 727.5 J B) Coefficient of Kinetic Friction = Force of Friction / Normal Force Because the surface is horizontal, the normal force = force of gravity 48(9.8) = 470.4 N You know the force of friction must equal 150 N because the crate is moving at a constant speed (thus, the net force = 0). coefficient = 150/470.4 = .3188
Sally applies a total force of 323 N with a rope to drag a wooden crate of mass 100 kg across a floor with a co?
The horizontal component of the applied force is: Fh = F × cos(θ) Fh = (323 N) × cos(45) Fh = 228.4 N That force should cause an acceleration of: Fh = m × a (228.4 N) = (100 kg) × a a = 2.284 m/s² The actual acceleration is only 0.1 m/s², which means that friction causes a deceleration of: f = (0.1 m/s²) - (2.284 m/s²) f = -2.184 m/s² Now you can find the force of friction. Ff = m × f Ff = (100 kg) × (-2.184 m/s²) Ff = -218.4 N (negative sign indicating that friction is acting against the direction of motion) The work done by friction is: Wf = Ff × d Wf = (-218.4 N) × (59.7 m) Wf = -13038 J
A block of 10 kg is pulled with a constant speed on a rough horizontal surface by a force of 19.6 N. What will be its coefficient?
0.2Since Normal Reaction force=10*9.8=98NFriction =x * 98 = 19.6x= coefficient of friction = 0.2
If a block of mass (m) is kept on a horizontal surface with friction coefficient [math]\mu[/math], then what would be the minimum force needed to move that block on the surface?
This is actually a tricky one since many people will end up gettingminimum force as [math]f = \mu.mg [/math]However Just read the question again , it says minimum force (There is no mention of minimum horizontal force)Now as we know Friction [math]force \sim Normal[/math]i.e. : if we minimize the normal reaction we can minimize the friction at the same time we also need to provide enough Horizontal force to ensure block gets a horizontal velocitySomething like shown in figure !Now the task is to find optimal angle “x” and Magnitude FApplying simple physics : (Assuming Normal reaction from surface is N)Balancing Vertical force[math]Fsinx+N = mg [/math] —>(1)[math]N = mg - Fsinx[/math]Balancing Horizontal force[math]Fcosx = Friction[/math]=> [math]Fcosx = \mu.N[/math]=> [math]Fcosx = \mu.(mg-Fsinx) [/math]=> [math]F(cosx+\mu.sinx) = \mu.mg [/math]=> [math]F = \frac{\mu.mg}{cosx+\mu.sinx} [/math] ==>(2)Now to minimize F(2) we need to maximize denominator of 2[math]max[/math]{[math]cosx+\mu.sinx[/math]}[math] = \sqrt{1+(\mu)^2} [/math] Basic TrigoSo minimum force needed would be[math]F = \frac{\mu.mg}{\sqrt{1+(\mu)^2}}[/math]