 Laveeza S.

# Greg places a bag of groceries on a table that is sloped at an angle of 15° to the horizontal. The box has a mass of 10 kg.

a. If there is no friction, what is the acceleration of the bag down the incline? (5 points)

b. If the coefficient of kinetic friction between the bag and the table is 0.18, what is the acceleration of the bag down the incline? (5 points)

c. What is the minimum coefficient of static friction between the bag and the table that will prevent the bag from slipping? (5 points)

By: Dominic S.

tutor
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.
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01/02/20 Dominic S.

tutor
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!
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01/02/20

Laveeza S.

thankyou, thankyou, thankyou very much.
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01/02/20

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