wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A rod of length 20 cm is kept along x axis. Two forces 5 N and 10 N are applied at distances 10 cm and 15 cm from point A as shown in figure. Find the point of application of net force from point A.


A
20 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
12.5 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
15 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
17.5 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 20 cm
Given, length of rod (l) = 20 cm
Position vector of point of application of F1=5 N(^j)
¯r1=10 cm ^i
Position vector of point of application of F2=10 N(^j)
¯r2=15 cm ^i
Torque due to F1
τ1=r1×F1=50×102 Nm(^k)
Torque due to F2
τ2=r2×F2=150×102Nm(^k)

Net force acting on rod (Fnet)=F1+F2=5 N(^j)
Since, r1×F1+r2×F2=r×Fnet we get
150×10250×102=5rr=20 cm

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon