Answer:
(a) The weight on the two nearer wheels combined is 323.73 N
(b) The weight on the two farther wheels combined is 824.04 N
Step-by-step explanation:
Here, we have
Taken the location of the four wheels as equidistant
The sum of moment at a point is equal to 0, that is ∑M = 0
Therefore, taking moment about the center of the cart with clockwise moments as positive and anticlockwise moments as negative, we have
2×Rw × 0.6 m + 102 × 0.3 × 9.81 - 2×Fw × 0.6 = 0
Where:
Rw = Rear wheel
Fw = Front wheel
Therefore, we have
1.2·Rw +300.186 = 1.2·Fw
1.2 Fw - 1.2 Rw = 300.186 N ....(1)
Also, ∑F = 0
That is 2·Rw + 2·Fw - 102×9.81 - 15 × 9.81 = 0
2·Rw + 2·Fw = 1,147.77 N............(2)
Placing Rw = (1.2 Fw - 300.186)/1.2 into equation 2 and solving, we obtain
Solving equations (1) and (2), we get
Rw = 161.865 N and Fw = 412.02 N
Therefore,
(a) The weight on the two nearer wheels is the weight on the Rw wheels given by
= 2 × 161.865 N = 323.73 N
(b) The weight on the farther wheels is the weight on the Fw wheels given by
= 2 × 412.02 N = 824.04 N.