Question

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question 18
1 pts
a 89 kg crate is suspended from the ceiling by two cables. each cable makes an angle of 33 degrees from the ceiling. calculate the full tension force along one of the cables in newtons.

Explanation:

Step1: Find the weight of the crate

The weight \( W \) of the crate is given by \( W = mg \), where \( m = 89\space kg \) and \( g = 9.8\space m/s^2 \). So \( W = 89\times9.8 = 872.2\space N \).

Step2: Analyze the vertical forces

Let \( T \) be the tension in each cable. The vertical component of each tension is \( T\sin(33^\circ) \). Since there are two cables, the sum of the vertical components balances the weight: \( 2T\sin(33^\circ)=W \).

Step3: Solve for \( T \)

Rearrange the formula: \( T=\frac{W}{2\sin(33^\circ)} \). Substitute \( W = 872.2\space N \) and \( \sin(33^\circ)\approx0.5446 \): \( T=\frac{872.2}{2\times0.5446}\approx\frac{872.2}{1.0892}\approx800.8\space N \) (rounded to a reasonable decimal place).

Answer:

Approximately \( 801\space N \) (or more precise value depending on calculation, e.g., if using more precise \( \sin(33^\circ) \) value like \( \sin(33^\circ)\approx0.544639035 \), \( T=\frac{89\times9.8}{2\times0.544639035}=\frac{872.2}{1.08927807}\approx800.7\space N \), so around \( 801\space N \))