Two forces each numerically equal to 10 dynes are acting as shown in the figure then tha resultant is :
The resultant will be a vector. So it is a tail of force on top and end on the head of the horizontal force. It is clear that the two forces and the resultant will form an equilateral triangle. So the resultant will be 10 dyne force acting on a line that makes 60 degree downwards angle with the horizontal.
If two forces are placed on adjacent sides of a parallelogram, the angle between them would be 120 degrees.
Now using the law of parallelogram we get
R = √10^2 + 10^2 + 2 x 10 x 10 x cos 120
R = √ 100 (cos 120 = - 1/2)
R = 10
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Answer laterfor a body of mass 50kg the velocity time graph is shown in fig. the force acting on the body is
Greetings!
Slope under (v-t) curve gives acceleration but since you haven't provided the diagram so if I assume
a = 0.5m/s
Then,Force = m a
=50 *0.5=25N.
thankyou
Good luck
two forces f1=2i-5j-6k and f2=-i+2j-k are acting on a boby at points (1,1,0) and (0,1,2) respectively. Find torque acting on body about point (-1,0,1)
Hey!There
First find torque due to force F1 about required point..And then torque due to force F2 about same point.Then take vector addition of these two torque..
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