Question

A tractor loader of mass $m_A$ is moving upward a hill with coals. The mass center (MC) of the tractor loader is located at G and the MC of the coal load in the loader bucket located at $G_L$ (see arrows in Fig. 3). The Coal Load is $m_B$ and its MC ($G_L$) is at height h = 3.1m and x = 1.5 m from the centre of front wheel M as shown in the figure.\ $m_A = 12.250Mg$\ $m_B = 2.381Mg$\ 1. Calculate the reactions by the ground on the wheels at M and N when the tractor is moving at an acceleration of 0.5 m/s² from rest.\ 2. Determine the maximum acceleration developed by the tractor loader to move forward on the hill without tipping over.\ 3. Justify the motion of the tractor loader case if the hill slope is 37º to the horizontal direction. The static friction between the wheels and the ground is 0.39.\ Note: 1Mg (Megagram) = 1000 Kg

          A tractor loader of mass $m_A$ is moving upward a hill with coals. The mass center (MC) of the tractor loader is located at G and the MC of the coal load in the loader bucket located at $G_L$ (see arrows in Fig. 3). The Coal Load is $m_B$ and its MC ($G_L$) is at height h = 3.1m and x = 1.5 m from the centre of front wheel M as shown in the figure.\
$m_A = 12.250Mg$\
$m_B = 2.381Mg$\
1. Calculate the reactions by the ground on the wheels at M and N when the tractor is moving at an acceleration of 0.5 m/s² from rest.\
2. Determine the maximum acceleration developed by the tractor loader to move forward on the hill without tipping over.\
3. Justify the motion of the tractor loader case if the hill slope is 37º to the horizontal direction. The static friction between the wheels and the ground is 0.39.\
Note: 1Mg (Megagram) = 1000 Kg

A tractor loader of mass mA is moving upward a hill with coals. The mass center (MC) of the tractor loader is located at G and the MC of the coal load in the loader bucket located at GL (see arrows in Fig. 3). The Coal Load is mB and its MC (GL) is at height h = 3.1m and x = 1.5 m from the centre of front wheel M as shown in the figure.mA = 12.250MgmB = 2.381Mg1. Calculate the reactions by the ground on the wheels at M and N when the tractor is moving at an acceleration of 0.5 m/s² from rest.2. Determine the maximum acceleration developed by the tractor loader to move forward on the hill without tipping over.3. Justify the motion of the tractor loader case if the hill slope is 37º to the horizontal direction. The static friction between the wheels and the ground is 0.39.Note: 1Mg (Megagram) = 1000 Kg

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