Home
Class 11
PHYSICS
When a force F acs on a body of mass m t...

When a force F acs on a body of mass m the acceleration product in the body is a . If htree equal forces `F_(1)=F_(2)=F_(3)=F` act on the same body as shown in figure the accleration produced is

A

`(sqrt(2)-1)`a

B

`(sqrt(2)+1)`a

C

`sqrt(2)a`

D

a

Text Solution

Verified by Experts

The correct Answer is:
A
(a) F=ma
Resultant of three forces `F_(1),F_(2)" and " F_(3)` will be `(sqrt(2)-1)`F.
Therefore, acceleration of body is also `(sqrt(2)-1)a`
Promotional Banner

Topper's Solved these Questions

  • LAWS OF MOTION

    DC PANDEY ENGLISH|Exercise Check point 5.3|20 Videos

  • LAWS OF MOTION

    DC PANDEY ENGLISH|Exercise Check point 5.4|20 Videos

  • LAWS OF MOTION

    DC PANDEY ENGLISH|Exercise Check point 5.1|20 Videos

  • KINEMATICS 1

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|15 Videos

  • LAWS OF THERMODYNAMICS

    DC PANDEY ENGLISH|Exercise Level 2 Subjective|18 Videos

Similar Questions

Explore conceptually related problems

When a force F acts on a body of mass m the acceleration product in the body is a . If three equal forces F_(1)=F_(2)=F_(3)=F act on the same body as shown in figure the acceleration produced is

Two forces F_(1) " and " F_(2) are acting on a rod abc as shown in figure.

When a force F acts on a particle of mass m, the acceleration of particle becomes a. now if two forces of magnitude 3F and 4F acts on the particle simultaneously as shown in figure, then the acceleration of the particle is

A force produces an acceleration of a_(1) in a body and the same force produces an acceleration of a_(2) in another body. If the two bodies are combined and the same force is applied on the combination, the acceleration produced in it is

A force F is acted on a body as shown in the figure. Find the impulse of the force in first four seconds.

A force produces an acceleration of 4ms^(-2) in a body of mass m_(1) and the same force produces an acceleration of 6ms^(-2) in another body oa mass m_(2) . If the same force is applied to (m_(1)+m_(2)) , then the acceleration will be

If force F is applied on a body and it moves with a velocity v, the power will be

A body of mass 5 kg is at rest. Three force F_(1) = 10 N due North F_2=10N along East and F_(3) = 10 sqrt(2) N along N-W act on it simultaneously. The acceleration produced in the body is,

A 10 N force is applied on a body produces an acceleration of 1 m//s^(2) . The mass of the body is

A rod of length L and mass M is acted on by two unequal force F_(1) and F_(2) ( lt F_(1)) as shown in the following figure. The tension in the rod at a distance y from the end a is given by