If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero. So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".
If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.
And if there are four ropes holding it up, then the sum of
the four tensions is 100N. If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.
<span>a+b= ?
3i +3j + (3i -3j) = ?
3i + 3j + 3i -3j =?
= 6i + 0j</span>
Answer:
naol may pinoy dito nag jakoI
<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>