A car is stuck in the mud. A tow truck pulls on the car with the arrangement sho

Question

A car is stuck in the mud. A tow truck pulls on the car with the arrangement shown in the figure below. The tow cable is under a tension of of 2,920 N and pulls downward and to the left on the pin at its upper end. The light pin is held in equilibrium by forces exerted by the two bars A and B. Each bar is a strut; that is, each is a bar whose weight is small compared to the forces it exerts and which exerts forces only through hinge pins at its ends. Each strut exerts a force directed parallel to its length. Determine the force of tension or compression in each strut. Proceed as follows. Make a guess as to which way (pushing or pulling) each force acts on the top pin. Draw a free-body diagram of the pin. Use the condition for equilibrium of the pin to translate the free-body diagram into equations. From the equations calculate the forces exerted by struts A and B. If you obtain a positive answer, you correctly guessed the direction of the force. A negative answer means the direction should be reversed, but the absolute value correctly gives the magnitude of the force. If a strut pulls on a pin, it is in tension. If it pushes, the strut is in compression. Identify whether each strut is in tension or in compression. (Assume theta 1 = 57 degree and theta 2 = 51 degree.) force exerted by strut A force exerted by strut B

Explanation / Answer

Now the tension on the tow cable is 2920 N
Hence
We can write this as
Tc= - 2920sin57i - 2920cos57j

let the force on the bar a be Taj

The force in the bar b be
bcos51i +bsin51j

Hence balancing the equation for horizontal components :

- 2920sin57 + bcos51 = 0
b = 3891.37 N

Now balancing for vertical we get
- 2920cos57 + a + 3891.37sin51 = 0
a = -1433.82 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.