0 تصويتات
175 مشاهدات
بواسطة
A uniform board weight 40 N supports two children weighing 500 N and 350 N. If the support is under the center of gravity of the board and, if the 500 N child is sitting 1.5 m from the center,
(1) determine upward force exerted on the board.
(2) determine where the 350 N child should sit to balance the system.

1 إجابة واحدة

0 تصويتات
بواسطة
(1) To find the upward force exerted on the board when the 500 N child is sitting 1.5 m from the center, we can use the principle of moments. The sum of the clockwise moments must equal the sum of the counterclockwise moments for the board to be in equilibrium.
Clockwise Moments = 500 N * 1.5m (due to the child)
Counterclockwise Moments = F (upward force) * 0.75 m (halfway between the children)
Since the board is in equilibrium, the two moments are equal:
500 N * 1.5 m = F * 0.75 m
750 N = 0.75 F
Dividing by 0.75:
F = 1000 N
Therefore, the upward force exerted on the board is 1000 N.
(2) To determine where the 350 N child should sit to balance the system, we can use the principle of moments again. We already know the clockwise moment due to the 500 N child:
500 N * 1.5 m = 750 Nm
The counterclockwise moment due to the 350 N child sitting at a distance x from the center would be:
350 N * x m
For the board to be in equilibrium, the sum of the clockwise moments must equal the sum of the counterclockwise moments:
750 Nm = 350 N * x m
Solving for x:
x = 750 Nm / 350 N
x = 2.14 m
Therefore, the 350 N child should sit at a distance of 2.14 m from the center to balance the system.

اسئلة متعلقة

حلول -------- حفظ.com
...