Answer:
Answer (a) 4.378 N-m
(b) 10.21 N-m
Explanation:
(a) FB= Frictional Force
For 30N force (p=30N)
P x (400+200) +(FB X50) -(NB X 200) =0
(30 X 600) +(0.3 X NB X 50) -(NB X 200) =0
NB= 97.29N
FB = 0.3 X 97.29 = 29.187N
Frictional torque on the wheel = FB x radius
= (29.189 x 0.15) N-m
=4.378 N-m which is less than 5 N-m
Since the torque is less than 5N-m the 30N force is not enough to hold the wheel without moving.
(b) For 70N force (p=70N)
P x (400+200) +(FB X 50) - (NB X 200) =0
(70 X 600) +(0.3 X NB X 50) -(NB X 200) =0
NB= 227.02N
FB = 0.3 X 227.02N = 68.10N
Frictional torque on the wheel = FB x radius
= (68.10 x 0.15) N-m
=10.21 N-m which is greater than 5 N-m
Since the torque(10.21 N-m) is greater than 5N-m the 70N force will be enough to hold the wheel at a place without moving.