Dominic S. answered • 01/02/20
Associates in Math and Science, experienced tutor
The box of groceries placed on an inclined plane can slide down it. The box will slide due to the gravity affecting it, but will be countered by other forces like friction or the normal force.
Part A:
Along the inclined plane, the gravitational force component parallel with the surface of the incline (the x axis) turns out to be
Fgx = m*g*sin(15)
after some vector trigonometry. The variable 'm' represents the objects mass in kg, 'g' represents the gravitational acceleration on earth, 9.8 m/s2, and the sine of the angle reduces the magnitude of the force proportionally to the angle it is off of the normal, in this case, 15 degrees. To find the acceleration down the plane, in the x direction, it is necessary to sum all forces in the x direction. Part A has us consider this situation in an environment where no friction exists between the inclined plane and the box, thus, there is no resisting force to the x component of the force of gravity, so the sum of forces in the x direction is
∑ Fx : m*g*sin(15) = m*a
setting it equal to 'm*a' allows us to interpret the scenario based on a certain mass, 10 kg. Substituting in known values and canceling out mass because it exists on each side of the equation gives us the new equation:
10*9.8*sin(15) = a
2.536 m/s2 = a, the mass' acceleration down the inclined plane.
Dominic S.
Part C: This part of the question has us evaluate the situation of a box not moving on the inclined plane, or when acceleration is 0. We return to our sum of forces in the x direction that we obtained in Part B, except we are now interested in the coefficient of static friction: ∑ Fx : m*g*sin(15) - m*g*cos(15)*µs = m*a and we want to find out what the value of 'µs' when acceleration is 0 so the equation m*g*sin(15) - m*g*cos(15)*µs = 0 becomes (m*g*sin(15)) / (m*g*cos(15)) = µs Which becomes tan(15) = µs 0.2679 = µs, the coefficient of static friction required to keep the box on the incline plane without slipping down it. Hope this helps!01/02/20
Laveeza S.
thankyou, thankyou, thankyou very much.01/02/20
Dominic S.
Part B: Part B is very similar to Part A, except this time it wants us to consider friction. What this changes most is the sum of forces in the x direction. Now, there is a force of friction (Ff) opposing the force of gravity in the x direction. The new sum of forces in the x direction equation is ∑ Fx : m*g*sin(15) - Ff = m*a all variables are known in this equation except 'Ff ' and 'a', but we're looking for 'a' so we must substitute Ff with some other known information. The equation Ff = Fn * µk will be useful to us here. µk is given, but Fn is still not known, but we can find it. Fn is the force applied to the underside of the box due to the surface contact. Because the box is in on the incline unsupported, the normal force is the only force opposing the force of gravity in the y direction and we know it is not accelerating off of the plane's face, so 'a = 0'. Analyzing forces in the y direction gives us the equation ∑ Fy : m*g*cos(15) - Fn = m * a which, after simplifying the right side, becomes ∑ Fy: m*g*cos(15) - Fn = 0 which becomes m*g*cos(15) = Fn, now giving us a known value for the normal force to plug into the force of friction equation. Ff = m*g*cos(15)*µk is now a useful equation to plug into the sum of forces in the x direction: ∑ Fx : m*g*sin(15) - Ff = m*a becomes ∑ Fx : m*g*sin(15) - m*g*cos(15)*µk = m*a where all values are known except 'a'. Solving for 'a' gives (m*g*sin(15) - m*g*cos(15)*µk) / m = a plugging in know values gives (10*9.8*sin(15) - 10*9.8*cos(15)*0.18) / 10 = a 0.832533 m/s2 = a, so with friction, the mass experiences a much lower acceleration.01/02/20