Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 1 (Q.No. 14)
14.
The primary unbalanced force is maximum when the angle of inclination of the crank with the line of stroke is
Discussion:
7 comments Page 1 of 1.
Prasanajit Das said:
3 years ago
Then, For a minimum?
Ashutosh said:
7 years ago
Yes, both (B&D) will be the correct answer coz F = MRW * 2Cos(Q).
if Q=0 then cos(Q)=1.
if Q=180 then cos(Q)=-1.
if Q=360 then cos(Q)=1.
So if we are considering the only magnitude then both answer will be correct.
if Q=0 then cos(Q)=1.
if Q=180 then cos(Q)=-1.
if Q=360 then cos(Q)=1.
So if we are considering the only magnitude then both answer will be correct.
(1)
Lok said:
7 years ago
Please explain it clearly.
Pradeep gk said:
8 years ago
Primery unbalanced force = mw^2*rcos(θ).
Rahul mandal said:
9 years ago
If we talk about magnitude. Then both b and d are true.
(1)
Kuldeep said:
9 years ago
Answer (b&d). Both are correct.
(1)
Sahib PEC said:
1 decade ago
Yes. (B) the reason being unbalanced force along the line of stroke would be mw^2*rcos(thita)and for thita= 0 and 180, cos of thita becomes max and hence F becomes max.
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