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calculating balanced forces a box at rest on a ramp is in equilibrium, …

Question

calculating balanced forces
a box at rest on a ramp is in equilibrium, as shown
what is the force of static friction acting on the box?
round your answer to the nearest whole number.
n
what is the normal force acting on the box? round
your answer to the nearest whole number.
n

Explanation:

Step1: Find Static Friction Force

The box is in equilibrium, so static friction \( F_{fs} \) balances the component of gravity along the ramp. The formula for the component of gravity along the ramp is \( F_{gx} = F_g \sin(\theta) \). Here, \( F_g = 735 \, \text{N} \) and \( \theta = 20^\circ \).

$$ F_{fs} = F_g \sin(20^\circ) = 735 \times \sin(20^\circ) $$
$$ \sin(20^\circ) \approx 0.3420, \quad F_{fs} \approx 735 \times 0.3420 \approx 251.37 \approx 251 \, \text{N} $$

Step2: Find Normal Force

The normal force \( F_N \) balances the component of gravity perpendicular to the ramp. The formula for this component is \( F_{gy} = F_g \cos(\theta) \).

$$ F_N = F_g \cos(20^\circ) = 735 \times \cos(20^\circ) $$
$$ \cos(20^\circ) \approx 0.9397, \quad F_N \approx 735 \times 0.9397 \approx 690.78 \approx 691 \, \text{N} $$

Answer:

Static friction force: \( \boldsymbol{251} \, \text{N} \)
Normal force: \( \boldsymbol{691} \, \text{N} \)