23.5 DEGREES is the answer.
well it depends on how much force was added but if they both have the same amount of force and they were rolled at the same time either they will bounce backwards or roll backwards not thats your question will they roll backwards are bounce backwards?
Answer:
Explanation:
Given that:
width b=100mm
depth h=150 mm
length L=2 m =200mm
point load P =500 N
Calculate moment of inertia

Point C is subjected to bending moment
Calculate the bending moment of point C
M = P x 1.5
= 500 x 1.5
= 750 N.m
M = 750 × 10³ N.mm
Calculate bending stress at point C

Calculate the first moment of area below point C

Now calculate shear stress at point C


Calculate the principal stress at point C
![\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa](https://tex.z-dn.net/?f=%5Csigma_%7B1%2C2%7D%3D%5Cfrac%7B%5Csigma_x%2B%5Csigma_y%7D%7B2%7D%20%5Cpm%5Csqrt%7B%28%5Cfrac%7B%5Csigma_x-%5Csigma_y%7D%7B2%7D%20%29%20%2B%20%28%5Ctau%29%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B666.67%2B0%7D%7B2%7D%20%5Cpm%5Csqrt%7B%28%5Cfrac%7B666.67-0%7D%7B2%7D%20%29%5E2%20%5Cpm%2844.44%29%5E2%7D%20%5C%20%5B%20%5Csigma_y%3D0%5D%5C%5C%5C%5C%3D333.33%5Cpm336.28%5C%5C%5C%5C%20%5Csigma_1%3D333.33%2B336.28%5C%5C%3D669.61KPa%5C%5C%5C%5C%5Csigma_2%3D333.33-336.28%5C%5C%3D-2.95KPa)
Calculate the maximum shear stress at piont C

Answer:
<u>Given</u><u> </u><u>-</u>
- Initial Velocity, u = 114 m/s
- Final velocity, v = 77 m/s.
- Time taken, t = 9 sec.
<u>To</u><u> </u><u>find</u><u> </u><u>-</u><u> </u>
<u>Solu</u><u>tion</u><u> </u><u>-</u>
Here, using the equation of motion v = u + at we can find the acceleration easily.
★ Here,
- V = Final velocity
- U = Initial Velocity
- A = Acceleration
- T = Time.
<u>Subs</u><u>tituting</u><u> </u><u>the</u><u> </u><u>values</u><u> </u><u>-</u>
→ 77 = 114 + a(9)
→ 9a = 114 - 77
→ 9a = 37
→ a = 37/9
→ a = 4.1 m/s
<u>There</u><u>fore</u><u>,</u><u> </u><u>the</u><u> </u><u>accele</u><u>ration</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>car</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>4</u><u>.</u><u>1</u><u> </u><u>m</u><u>/</u><u>s</u><u>.</u>