Kinematics: The crate lies on a plane tilted at an angle θ = 25 ∘ to the horizontal, with μk = 0.19.?
self-motive force Fm = m*g*sin 25° friction force Fr = m*g*cos 25°*μ accelerating force Fa = Fm-Fg = m*g*(sin 25°-cos 25°*μ) acceleration a = Fa/m = m*g*(sin 25°-cos 25°*μ)/m = g*(sin 25°-cos 25°*μ) = 0.25*g m/sec^2 incline lenght ℓ = h/sin 25 = 8.15/0.4226 = 19.28 m V = √2*a*ℓ = √2*0.25*9.8*19.28 = 9.72 m/sec or t = 2ℓ/a = √19.28*2*4/9.8 = 3.97 sec V = a*t = 3.97*9.8/4 = 9.72 m/sec or V = 2ℓ/t = 19.28*2/3.97 = 9.72 m/sec
A crate lies on the plane tilted at an angle 25° to the horizontal with uk=.19...?
A.) Determine the acceleration of the block as it slides down the plane. B.) If the block starts from rest 12.0 meters up the plane from its base, what will be the blocks speed when it reaches the bottom of the incline?
A crate lies on a plane tilted at an angle 25.0 degrees to the horizontal...?
This question requires you to draw a picture and all the forces. You problem looks something like this: http://ts2.mm.bing.net/th?id=HN.60805546... Instead of thinking horizontal and vertical, you need to rotate the coordinate system so that you are looking at the parallel and perpinduclar to the plane directions. Sum of the forces PARALLEL to the plane (up the plane is positive) = -m*a = -m*g*sin(theta) - friction Note: the reason for the -m*a is because we declared "up the plane is positive" and the block is accelerating down the plane. Also, the component of weight parallel to the ramp, m*g*sin(theta), acts down the plane and friction opposes motion, so since the block is accelerating down the plane, friction acts up the plane. Sum of the forces PERPENDICULAR to the plane (away from the plane is positive) = 0 = N - m*g*cos(theta) N = m*g*cos(theta) Note: Instead of m*a it is 0 is because the block is not accelerating off the plane. The only forces in the perpendicular direction are the normal force, which acts away from the plane (positive), and the componenet of weight perpendicular to the plane, m*g*cos(theta), which acts into the plane (negative). Definition of friction friction =u*N Sub in N friction = u*m*g*cos(theta) Sub that into the parallel equation and solve for a -m*a = -m*g*sin(theta) + u*m*g*cos(theta) All the m's cancel -a = -g*sin(thetA) + u*g*cos(theta) a = g*sin(theta) - u*g*cos(theta) Plug in numbers a = 10 m/s^2 * sin(25) - 0.19*10m/s^2 * cos(25) = 2.50 m/s^2 b) Now that you have the acceleration, use kinematics to find the speed vf^2 = vi^2 + 2*a*d With vi = 0 vf = sqrt(2*a*d) Plug in numbers vf = sqrt(2 * 2.50 m/s^2 * 8.15 m) = 6.39 m/s
Physics: The crate lies on a plane tilted at an angle that is 28 degrees to the horizontal?
The crate lies on a plane tilted at an angle that is 28 degrees to the horizontal, with the kinetic friction coefficient = 0.16. A) Determine the acceleration of the crate as it slides down the plane. B) If the crate starts from rest at height of 8.15 m from base of the plane, what will be the crate's speed when it reaches the bottom of the incline?
If the coefficient of friction of a plane inclined at 30 degrees is .4, what is the acceleration of the body sliding freely on it?
The forces acting on the body are (see figure).1.weight mg ,vertically downward .2. Normal force ,N perpendicular to the surface of incline.3. Frictional force , parallel (upward) to incline)Since we are interested in the motion parallel to the inclined surface, we take one component of mg parallel ( downward) to inclined plane. This component ismg sin(theta).Another component of mg perpendicular to inclined surface mg cos(theta) = N, because there is no motion perpendicular to the inclined surface.Now frictional force f=uN=umg cos (theta). Here, u is coefficient of friction.If a is acceleration of the body parallel to incline, the equation of motion ( Newton’s second Law of motion) will bema= mg sin (theta)- u mg cos (theta). m gets cancelled from both sides. Therefore .a=g sin (theta)-u g cos (theta).Put ( theta)= 30 degree , u=0.4 and g=9.8 m/s^2 in this equation. Then,a=(9.8)(1/2)-(0.4)(9.8)(root(3)/2). I hope now you will find value of a.
The carton shown in Fig. 4-55 lies on a plane tilted at an angle θ = 26.0° to the horizontal, with µk = 0.07.?
The carton shown in Fig. 4-55 lies on a plane tilted at an angle θ = 26.0° to the horizontal, with µk = 0.07. Figure 4-55 (a) Determine the acceleration of the carton as it slides down the plane. ________m/s2 (down the plane) (b) If the carton starts from rest 9.20 m up the plane from its base, what will be the carton's speed when it reaches the bottom of the incline? _________ m/s
A box is sitting on a ramp inclined at 30 degree angle. The ramp has a coefficient of static friction of 0.90. What is its mass?
Conclusion: the box will never move regardless of its mass since frictional force will always keep it stationary.
For a box sitting motionless on an inclined plane, the force of friction is equal to?
First draw a free body diagram of the block. A free body diagram shows all the forces acting on the object.Notice that I have defined a rotated set of axes and I labelled them x’ and y’. The x’-axis is parallel to the inclined plane and the y’-axis is perpendicular to the plane. I chose positive x’-axis down the plane. The component of the weight (mg) acting down the plane is found by resolving the weight into components as shown in my example below using [math]\theta=30[/math] :So the component of the weight acting down the plane is [math](mg)sin\theta[/math]. The friction force acts opposite the direction of motion (up the plane) as shown on my free body diagram. To determine [math]F_{fric}[/math] , write a static equilibrium equation in the x’ direction. Static equilibrium is a fancy way of saying all the forces balance:[math]\Sigma F_{x'}=0[/math][math](mg)sin\theta-F_{fric}=0[/math][math]\therefore[/math] [math]F_{fric}=(mg)sin\theta[/math]This result only applies if the angle of the inclined plane is less than the maximum possible angle when slippage occurs. So the force of friction will increase with increasing [math]\theta[/math] until the box begins to slip. If slippage is impending, then we have the maximum possible friction force given by:[math]F_{fric}=\mu_{s} F_{N}[/math]In this case, first find the normal force between the ramp and the object. To determine [math]F_{N}[/math] , write a static equilibrium equation in the y’ direction:[math]\Sigma F_{y'}=0[/math][math]F_{N}-(mg)cos\theta=0[/math][math]F_{N}=(mg)cos\theta[/math]Now substitute the normal force into:[math]F_{fric}=\mu_{s} F_{N}[/math][math]F_{fric}=\mu_{s} (mg)cos\theta[/math]
If a block slides down a 30 degree incline at a constant speed, what is the coefficient of dynamic friction between the block and the incline?
θ = 30 degreesvelocity is constant therefore acceleration a = 0Please refer to my analysis using the diagram below.The coefficient of dynamic friction is equal to 0.5773.