Answer:
The "normal force" on the "cart" 63.893 N .
Explanation:
To find normal force on the cart, use the equation
Normal force = mg + F sinx,
“m” being the object's mass,
“g” being the acceleration of gravity,
“x” being the angle of the cart
Given values
M = 7.33 kg
F = 14.7 N
[tex]x=-32.7^{\circ}[/tex]
[tex]\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }[/tex]
Substitute the values in above equation
Normal force = (7.33 × 9.8) + 14.7 sin(-32.7°)
Normal force = 71.834 + 14.7 × (-0.5402)
Normal force = 71.834 - 7.94094
Normal force = 63.893 N
The "normal force" on "the cart" 63.893 N .
The normal force on the cart is 79.7 N
Explanation:
In order to find the normal force, we have to analyze the forces acting on the cart on the vertical direction.
In the vertical direction, we have the following forces:
The weight of the cart, downward, of magnitude , where m is the mass of the cart and g is the acceleration of gravity
The normal force on the cart, upward, we indicate it with N
The component of the pushing force acting in the vertical direction, downward, of magnitude , where F is the magnitude of the force and is the angle of the force with the horizontal
Therefore, the equation of the forces on the cart in the vertical direction is:
where the net force is zero since the cart is balanced in the vertical direction. We have:
We take the angle as positive since we are already considering the downward direction in the equation.
Substituting and solving for N, we find the normal force:
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