Solving a Ramp Problem with 1kg Block

[Inquiring_Mike]

Hi, I have a ramp problem question that I don't think makes sense..
Question:
A 1kg block is placed on a ramp where the angle at which the slope is increased by 1 degree until it falls down the ramp. The block falls down the ramp when the angle is 25 degrees, find the a) force of gravity, b) normal force c) applied force, d) force of static friction and e) the coefficient of static friction... I know how to get the force of gravity, but none of the others... ?

 

[HallsofIvy]

The gravitational force is, of course, -mg.

The "normal force" is the force perpendicular to the surface. Set up a diagram with the gravitational force vector downward as the hypotenuse of a right triangle, one leg perpendicular to the surface and one leg parallel to it. You can show that the angle of that triangle at the surface is the same as the angle the surface makes with the horizontal: in this case 25 degrees: use that to show that the normal force is -mg cos(25). The force along the ramp is
-mg sin(25).

If the mass does not begin to move until the ramp is at a 25 degree angle, then the friction force at that angle must be exactly enough to offset the force along the ramp: the friction force is
mg sin(25).

The friction force is given by the "coefficient of friction" times the normal force. Since you know that the normal force is -mg cos(25) and the friction force is mg sin(25): the coefficient of friction must be -sin(25)/cos(25).

I don't see any "applied force" here.

 

[M98Ranger]

Hi there,

I completely understand your confusion with this problem. It can definitely be a bit tricky to figure out all the different forces at play. Let's break it down step by step.

First, let's start with the force of gravity. This is simply the weight of the block, which we can calculate using the formula Fg = mg, where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2 on Earth). So in this case, the force of gravity would be 1kg x 9.8 m/s^2 = 9.8 N.

Next, let's look at the normal force. This is the force exerted by the ramp on the block, perpendicular to the surface. In this case, since the ramp is at an angle, we need to take into account the component of the force that is perpendicular to the ramp. Using trigonometry, we can determine that the normal force is equal to the force of gravity multiplied by the cosine of the angle of the ramp. So in this case, the normal force would be 9.8 N x cos(25 degrees) = 8.86 N.

The applied force is the force that is being exerted on the block to push it up the ramp. In this problem, it is not given, so we cannot determine it.

Next, let's look at the force of static friction. This is the force that keeps the block from sliding down the ramp before it reaches the critical angle of 25 degrees. The formula for static friction is Ffs = μsFn, where μs is the coefficient of static friction and Fn is the normal force. Since we have already calculated the normal force, we just need to determine the coefficient of static friction. This can be done by setting up an equation using the fact that the block is just about to slide at 25 degrees. So we have Fg sin(25 degrees) = μsFn, which can be rearranged to solve for μs. In this case, μs = Fg sin(25 degrees)/Fn = (9.8 N x sin(25 degrees)) / 8.86 N = 0.45.

Finally, let's look at the applied force. Since the block is just about to slide, the applied force will be equal to the force of static friction, which we determined to be 0.45 x 8.

 

[M98Ranger]

Hi there, I understand your confusion with this ramp problem. Let's break it down and solve it step by step.

First, let's define the given information: a 1kg block, a ramp with a slope that increases by 1 degree, and the block falls down when the angle reaches 25 degrees.

a) Force of gravity: Since we know the mass of the block is 1kg, we can use the equation F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2). So the force of gravity acting on the block is 9.8N.

b) Normal force: The normal force is the force exerted by the surface on the object. In this case, it is the force of the ramp pushing up on the block to counteract the force of gravity. Since the block is not moving up or down the ramp, the normal force must be equal in magnitude to the force of gravity. So the normal force is also 9.8N.

c) Applied force: In this problem, there is no applied force mentioned. So we can assume that there is no external force acting on the block.

d) Force of static friction: The force of static friction is the force that opposes the motion of an object on a surface. In this case, since the block is not sliding down the ramp, the force of static friction must be equal in magnitude to the force of gravity. So the force of static friction is also 9.8N.

e) Coefficient of static friction: The coefficient of static friction is a value that represents the amount of friction between two surfaces. It is specific to the materials of the surfaces in contact. In this problem, we can use the equation F = μN, where μ is the coefficient of static friction and N is the normal force. Since we know the normal force is 9.8N, we can rearrange the equation to find the coefficient of static friction. So μ = F/N = 9.8N/9.8N = 1.

I hope this breakdown helps you understand how to solve this ramp problem. If you have any further questions, please don't hesitate to ask. Good luck!