Question

For the arm shown in the picture below, calculate the torque, and the find the “torque load” assuming a current limit of 2 amps. Use the motor specification given in the table and the formulas to solve the problem. Video how your solved the problem (using a smart phone) and submit it to this assignment.

Answer 1

The torque of a current-carrying loop is calculated using T = NIAB sin θ, and the maximum torque occurs when θ is 90 degrees, simplifying to T = NIAB.

Step-by-step explanation:

Calculating Torque on a Current-Carrying Loop

When calculating the torque on a current-carrying loop in a magnetic field, we use the formula T = NIAB sin θ, where T is the torque, N is the number of turns in the loop, I is the current, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the direction of the current and the magnetic field. To find the maximum torque, we use θ = 90 degrees (or π/2 radians), since sin(π/2) = 1, and thus, T = NIAB. Remember that torque has units of N·m which must equal the units derived from the given units of A·m²·T.

For example, to calculate the maximum torque on a 100-turn square loop of wire, 10.0 cm on a side, carrying 15.0 A of current in a 2.00-T magnetic field, you would first calculate the area A = 0.1 m * 0.1 m, then multiply by the number of turns, current, and magnetic field strength.

Answer 2

Explanation

the torque load is given by:

`\begin{gathered} Torque\text{ load=}\frac{(\text{current limit-fre}e\text{current)}\cdot\text{stall torque}}{(\text{stall current-}fre\text{ current)}} \\ \end{gathered}`

hence

Step 1

Torque load

a)Let

`\begin{gathered} free\text{ sp}eed=100\text{ RPM} \\ Stalltorque=1.67\text{ N}\cdot m \\ Stall\text{ current=}4.8\text{ Amps} \\ fre\text{ current=}0.37\text{ Amps} \\ \text{Current limit= 2 Amps} \end{gathered}`

b)now, replace in the formula

`\begin{gathered} Torque\text{ load=}\frac{(\text{current limit-fre}e\text{ current)}\cdot\text{stall torque}}{(\text{stall current-}fre\text{ current)}} \\ Torque\text{ load=}\frac{(2\text{ Amps-}0.37\text{ amps)}\cdot1.67\text{ N}\cdot m}{(4.8\text{ Amps-}0.37\text{ Amps)}} \\ Torque\text{ load=}\frac{(1.63\text{amps)}\cdot1.67\text{ N}\cdot m}{(4.43\text{ Amps)}} \\ Torque\text{ load=0.61 N}\cdot m \end{gathered}`

therefore, the answer is

torque load: 0.61 N*m

Step 2

Torque:

the torque is given by the expression

`\text{Torque}=\text{ force}\cdot\text{ distance}`

so , after checking the imagen

let

distance=0.5 m

force=30 N

now,replace and calculate

`\begin{gathered} \text{Torque}=\text{ force}\cdot\text{ distance} \\ \text{Torque = 30 N}\cdot0.5\text{ m} \\ \text{Torque =}15\text{ N}\cdot m \end{gathered}`

so

torque: 15 N*m

I hope this helps you