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

A body of weight 50 N is dragged slowly with constant velocity on a horizontal surface as shown in the figure, with a force of F=28.2 N at an angle of 45 with horizontal. The net contact force acting on the body will be


A
1017 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1019 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1013 N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1012 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1013 N
From the free body diagram of the block


It is given in the problem that the body moves with constant velocity hence acceleration, a=0 an so the frictional force will be equal to the force applied in horizontal direction
f=Fcos45o=28.22=20 N where (2 1.41)

Now for calculating for the normal force, we balance vertical forces
N+Fsin45o=50N=5020=30 N

Hence the net contact force acting on the body will be
f2+N2=202+302=1013 N

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Rubbing It In: The Basics of Friction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon