Answer:
a) a = 0.477 m/s^2
b) u = 0.04862
Explanation:
Given:-
- The rotational speed of the turntable N = 33 rev/min
- The watermelon seed is r = 4.0 cm away from axis of rotation.
Find:-
(a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip
Solution:-
- First determine the angular speed (w) of the turntable.
w = 2π*N / 60
w = 2π*33 / 60
w = 3.456 rad/s
- The watermelon seed undergoes a centripetal acceleration ( α ) defined by:
α = w^2 * r
α = 3.456^2 * 0.04
α = 0.477 m / s^2
- The minimum friction force (Ff) is proportional to the contact force of the seed.
- The weight (W) of the seed with mass m acts downwards. The contact force (N) can be determined from static condition of seed in vertical direction.
N - W = 0
N = W = m*g
- The friction force of the (Ff) is directed towards the center of axis of rotation, while the centripetal force acts in opposite direction. The frictional force Ff = u*N = u*m*g must be enough to match the centripetal force exerted by the turntable on the seed.
Ff = m*a
u*m*g = m*a
u = a / g
u = 0.477 / 9.81
u = 0.04862
Answer:
Fa=774 N
Fb=346 N
Explanation:
We will solve this problem by equating forces on each axis.
- On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
- On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative
While towing we know that car is mot moving in y-direction so net force in y-axis must be zero
⇒∑Fy=0
⇒
⇒
⇒
Given that resultant force on car is 950N in positive x-direction
⇒∑Fx=950
⇒
⇒
⇒
⇒
⇒
⇒ 
⇒
Therefore approximately, Fa=774 N and Fb=346 N
Answer:
See the explanation below
Explanation:
We must perform a sum of forces on the body, that sum of forces is equal to zero. That is, the body is in balance and does not move.
ΣF = 0
3 - 3 = 0
This force is negative and acts by pointing downwards.
