Let the center of the wheel be O, and let the endpoint of the handle as shown in the original picture be H.
Check the first picture attached. OH=48 and m(OHB)=20°.
|OB|=|OH|*sin20°=48 in * 0.342= 16.4 in
The handle of the handcart when lifted, moves around a circle formed, with center and radius |OH|=48 in.
To elevate the handle at 48 in off the ground, the handle first comes at point H', such that OH'//BH, then comes to point H" such that the distance of point H" to the ground is 48 inches. (check picture 1)
check picture 2, the distance of the point H'' to the line segment OH' is 48-16.4= 31.6 (inches)
31.6 = 48 * sinα
sin α=31.6/48=0.6583
so α = arcsin(0.6583)=41.17°
similarly, instead of α, 20° is the greatest angle such that the content dont spill out, and let h be the height above the line segment OH'.
h= 48*sin20=48*0.342=16.4
so the maximal height at which the content dont spill is 16.4+16.4=32.8 (inches)
Answer: 1. the contents would spill out, 2. maximal height: 32.8 inches
Check the first picture attached. OH=48 and m(OHB)=20°.
|OB|=|OH|*sin20°=48 in * 0.342= 16.4 in
The handle of the handcart when lifted, moves around a circle formed, with center and radius |OH|=48 in.
To elevate the handle at 48 in off the ground, the handle first comes at point H', such that OH'//BH, then comes to point H" such that the distance of point H" to the ground is 48 inches. (check picture 1)
check picture 2, the distance of the point H'' to the line segment OH' is 48-16.4= 31.6 (inches)
31.6 = 48 * sinα
sin α=31.6/48=0.6583
so α = arcsin(0.6583)=41.17°
similarly, instead of α, 20° is the greatest angle such that the content dont spill out, and let h be the height above the line segment OH'.
h= 48*sin20=48*0.342=16.4
so the maximal height at which the content dont spill is 16.4+16.4=32.8 (inches)
Answer: 1. the contents would spill out, 2. maximal height: 32.8 inches
