Question

quiz instructions
you can complete this review up to 3 times. it will be worth 2.5 points extra credit onto the exam.
question 18
1 pts
a 65 kg crate is suspended from the ceiling by two cables. each cable makes an angle of 56 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 \( m = 65\space kg \) and \( g = 9.8\space m/s^2 \). So \( W=65\times9.8 = 637\space N \).

Step2: Analyze vertical forces

Let \( T \) be the tension in each cable. The vertical component of each tension is \( T\sin\theta \), where \( \theta = 56^\circ \). Since there are two cables, the sum of vertical components balances the weight: \( 2T\sin\theta=W \).

Step3: Solve for tension \( T \)

Rearrange the formula: \( T=\frac{W}{2\sin\theta} \). Substitute \( W = 637\space N \) and \( \theta = 56^\circ \) ( \( \sin56^\circ\approx0.8290 \) ): \( T=\frac{637}{2\times0.8290}\approx\frac{637}{1.658}\approx384.2\space N \).

Answer:

\( \approx 384\space N \) (or more precisely \( \approx 384.2\space N \))