Question

1. the sign is in equilibrium and weighs 250.0n. θ = 35.0° a) draw a free body diagram b) calculate t₁ and t₂.

Explanation:

Step1: Analyze horizontal force equilibrium

In the horizontal direction, \(T_{1}\cos\theta=T_{2}\cos\theta\), so \(T_{1} = T_{2}\)

Step2: Analyze vertical force equilibrium

In the vertical direction, \(T_{1}\sin\theta+T_{2}\sin\theta = W\). Since \(T_{1} = T_{2}\), we can substitute \(T_{2}\) with \(T_{1}\), then \(2T_{1}\sin\theta=W\)

Step3: Solve for \(T_{1}\) and \(T_{2}\)

We know \(W = 250.0N\) and \(\theta=35.0^{\circ}\). From \(2T_{1}\sin\theta=W\), we can solve for \(T_{1}\):

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Using \(\sin(35.0^{\circ})\approx0.574\), we have \(T_{1}=\frac{250.0}{2\times0.574}\approx218N\). Since \(T_{1} = T_{2}\), \(T_{2}\approx218N\)

Answer:

\(T_{1}\approx218N\), \(T_{2}\approx218N\)